**1.** A man has to go $50$ m due north, $40$ m due east and
$20$ m due south to reach a field. (a) What distance he
has to walk to reach the field ? (b) What is his
displacement from his house to the field ?

1 None

Solutionsa) Distance travelled = $50$ + $40$ + $20$ = $110$ m

b) $AF$ = $AB$ - $BF$ = $AB$ - $DC$ = $50$ - $20$ = $30$ M $\\$ His displacement is $AD$

$AD$ = $\sqrt{AF^2-DF^2}$ = $\sqrt{30^2+40^2}$ = $50$ M

In $\Delta{AED}$ tan$\theta$ = $\frac{DE}{AE}$ = $\frac{30}{40}$ = $\frac{3}{4}$

$\Rightarrow$ $\theta$ = $tan^{-1}$ ($\frac{3}{4}$)

His displacement from his house to the field is $50$ m , $\\$ $tan^{-1}$($\frac{3}{4}$) north to east

**2.** A particle starts from the origin, goes along the X-axis
to the point $(20 m, 0)$ and then returns along the same
line to the point $(-20 m, 0)$. Find the distance and
displacement of the particle during the trip.

2 None

SolutionsO $\rightarrow$ Starting point origin. $\\$ i ) Distance travelled = $20$ + $20$ + $20$ = $60$ m $\\$ ii ) Displacement is only OB = $20$ m in the negative direction. $\\$ Displacement $\rightarrow$ Distance between final and initial position.

**3.** It is $260$ km from Patna to Ranchi by air and $320$ km
by road. An aeroplane takes $30$ minutes to go from Patna
to Ranchi whereas a delux bus takes $8$ hours. $\\$(a) Find
the average speed of the plane. $\\$(b) Find the average
speed of the bus.
$\\$ (c) Find the average velocity of the
plane. $\\$(d) Find the average velocity of the bus.

3 None

Solutionsa) $\ V_{avg}$ of plane ($\frac{Distance}{Time}$) = $\frac{260}{0.5}$ = 520$\frac{km}{hr}$

b) $\ V_{avg}$ of bus = $\frac{320}{8}$ = 40$\frac{km}{hr}$

c) Plane goes in straight path $\\$ velocity = $\vec{V}_{avg}$ = $\frac{260}{0.5}$ = 520$\frac{km}{hr}$

d) Straight path distance between plane to ranchi is equal to the displacement of bus.$\\$ $\therefore$ velocity = $\vec{V}_{avg}$ = $\frac{260}{8}$ = 32.5$\frac{km}{hr}$

**4.** When a person leaves his home for sightseeing by his
car, the meter reads $12352$ km. When he returns home
after two hours the reading is $12416$ km.$\\$ (a) What is the
average speed of the car during this period ? $\\$(b) What
is the average velocity ?

4 None

Solutionsa ) Total distance covered $12416$ - $12352$ = $64$ km in $2$ hours. $\\$ speed = $\frac{64}{2}$ = 32$\frac{km}{hr}$

b) As he returns to his house , the displacement is zero. $\\$ Velocity = $\frac{displacement}{time}$ = 0(zero)

**5.** An athelete takes $2.0$ s to reach his maximum speed of
$18.0$ km/h. What is the magnitude of his average
acceleration ?

5 None

SolutionsInitial velocity u = 0 ( $\therefore$ starts from rest) $\\$ final velocity v = 18 $\frac{km}{hr}$ = 5 sec.

(i.e max velocity) $\\$ The interval t = 2 sec $\\$ $\therefore$ acceleration = $a_{avg}$ = $\frac{v-u}{t}$ = $\frac{5}{2}$ = 2.5$\frac{m}{s^2}$

**6.** The speed of a car as a function of time is shown in
figure (3-E1). Find the distance travelled by the car in
$8$ seconds and its acceleration.

6 None

SolutionsIn the interval 8 sec the velocity changes from 0 to 20 $\frac{m}{s}$ $\\$ average acceleration = $\frac{20}{8}$ = 2.5 $\frac{m}{s^2}$ ($\frac{change in velocity}{time}$)

**7.** The acceleration of a cart started at $t = 0$, varies with
time as shown in figure (3-E2). Find the distance
travelled in $30$ seconds and draw the position-time graph.

7 None

SolutionsIn $1^{st}$ 10 sec $S_{1}$ = ut + $\frac{1}{2}$a$t^{2}$ $\Rightarrow$ 0 + ( $\frac{1}{2} \times 5 \times 10^2$) = 250 ft.

At 10 sec v = u + at = 0 + 5 $\times$ 10 = 50 $\frac{ft}{sec}$ .

$\therefore$ From 10 to 20 sec ($\Delta$t = 20 - 10 = 10 sec) it moves with uniform velocity 50 $\frac{ft}{sec}$ .

Distance $S_2 = 50 \times 10 = 500 ft.$

Between 20 sec to 30 sec Acceleration is constant i.e -5$\frac{ft}{s^2}$ . At 20 sec velocity is 50$\frac{ft}{sec}$.

t = 30 - 20 = 10 sec.

$S_3 = ut + \frac{1}{2}at^2 = 50 \times 10 + (\frac{1}{2})(-5)(10)^2$ = 250 m

Total distance travelled in 30 sec = $S_1$ + $S_2$ + $S_3$ = 250 + 500 + 250 = 1000 ft.

**8.** Figure $(3-E3)$ shows the graph of velocity versus time
for a particle going along the X-axis. Find $\\$ (a) the acceleration, $\\$ (b) the distance travelled in 0 to 10 s and $\\$
(c) the displacement in 0 to 10 s.

8 None

Solutionsa) Initial velocity u = 2 $\frac{m}{s}$. $\\$ Final velocity v = 8 $\frac{m}{s}$. $\\$ time = 10 sec, $\\$ acceleration = $\frac{v-u}{at}$ = $\frac{8-2}{10}$ = 0.6 $\frac{m}{s^2}$

b) $v^2$ - $u^2$ = 2aS $\\$ $\Rightarrow$ Distance S = $\frac{v^2-u^2}{2a}$ = $\frac{8^2-2^2}{2 \times 0.6}$ = 50 m.

c) Displacement is same as distance travelled .

Displacement = 50 m.

**9.** Figure (3-E4) shows the graph of the x-coordinate of a
particle going along the X-axis as a function of time.
Find $\\$ (a) the average velocity during $0$ to $10$ s, $\\$
(b) instantaneous velocity at $2$, $5$, $8$ and $12s$.

9 None

SolutionsDisplacement in 0 to 10 sec is 1000 m.

time = 10 sec.

$v_{avg}$ = $\frac{s}{t}$ = $\frac{100}{10}$ = 10 $\frac{m}{s}$

b) At 2 sec it is moving with uniform velocity $\frac{50}{2.5}$ = 20 $\frac{m}{s}$.

At 2 sec . $V_{inst}$ = 20 $\frac{m}{s}.$

At 5 sec it is at rest.

$V_{inst}$ = $zero$

At 8 sec it is moving with uniform velocity 20 $\frac{m}{s}$

$V_{inst}$ = 20 $\frac{m}{s}$

At 12 sec velocity is negative as it move toward initial position . $V_{inst}$ = -20 $\frac{m}{s}$

**10.** From the velocity—time plot shown in figure (3-E5), find
the distance travelled by the particle during the first $40$ seconds. Also find the average velocity during this
period.

10 None

SolutionsDistance in first 40 sec is , $\Delta{OAB}$ + $\Delta{BCD}$

= $\frac{1}{2} \times 5 \times 20$ + $\frac{1}{2} \times 5 \times 20$ = 100 m.

Average velocity is 0 as the displacement is zero.

**11.** Figure (3-E6) shows $x-t$ graph of a particle. Find the
time $t$ such that the average velocity of the particle
during the period $0$ to $t$ is zero.

11 None

SolutionsConsider the point B , at t = 12 sec $\\$ At t = 0 ; s = 20 m $\\$ and t = 12 sec ; s = 20m $\\$ So for time interval 0 to 12 sec. $\\$ Change in displacement is zero. $\\$ So average velocity = $\frac{displacement}{time}$ = 0 $\\$ $\therefore$ The time is 12 sec.

**12.** A particle starts from a point A and travels along the
solid curve shown in figure $(3-E7)$. Find approximately
the position B of the particle such that the average
velocity between the positions A and B has the same
direction as the instantaneous velocity at B.

12 None

SolutionsAt position B instantaneous velocity has direction along $\vec{BC}$. For average velocity between A and B.

$V_{avg}$ = $\frac{displacement}{time}$ = $\frac{(\vec{AB})}{t}$ $\\$ t= time

We can see that $\vec{AB}$ is along $\vec{BC}$ i.e they are in same direction.

The point is B (5m , 3m).

**13.** An object having a velocity $4.0$ m/s is accelerated at the
rate of $1.2$ m/s2 for $5.0$ s. Find the distance travelled
during the period of acceleration.

13 None

Solutionsu = 4 $\frac{m}{s}$ , a = 1.2 $\frac{m}{s^2}$ , t = 5 sec $\\$ Distance = s = $ut$ + $\frac{1}{2}at^2$

= $4(5) + \frac{1}{2}(1.2)5^2$ = $35$ m.

**14.** A person travelling at $43.2$ km/h applies the brake giving
a deceleration of $6.0$ $m/s2$ to his scooter. How far will it
travel before stopping ?

14 None

SolutionsInitial velocity u = $43.2$ $\frac{km}{hr}$ = $12$ $\frac{m}{s}$ $\\$ u = $12$ $\frac{m}{s}$ , v = 0 $\\$ a = $-6$ $\frac{m}{s^2}$ (deceleration)

Distance S = $\frac{v^2-u^2}{2(-6)}$ = $12$ m

**15.** A train starts from rest and moves with a constant
acceleration of $2.0$ $m/s2$ for half a minute. The brakes
are then applied and the train comes to rest in one
minute. Find $\\$(a) the total distance moved by the train,$\\$
(b) the maximum speed attained by the train and $\\$ (c) the
position(s) of the train at half the maximum speed

15 None

Solutions$Initial$ $velocity$ $u = 0$ $\\$ $Acceleration$ $a = 2$ $\frac{m}{s^2}$. $Let$ $final$ $velocity$ $be$ v ( before applying breaks ) $\\$ t = $30$ $sec$ $\\$ $v = u + at \Rightarrow 0 + 2 \times 30 = 60$ $\frac{m}{s}$

a) $S_1 = ut + \frac{1}{2}at^2 = 900 m$

when breaks are applied u' = $60 \frac{m}{s}$

$v' = 0 , t = 60$ sec ( 1 min)

Declaration $a'$ = $\frac{(v-u)}{t}$ == $\frac{(0-60)}{60}$ = -1 $\frac{m}{s^2}$. $\\$ $S_2$ = $\frac{(v')^2-(u')^2}{2a'}$ = $1800$ m $\\$ Total S = $S_1$ + $S_2$ = $1800$ + $900$ = $2700$ m = $2.7$ km.

b) The maximum speed attained by train v = $60$ $\frac{m}{s}$

c) Half the maximum speed = $\frac{60}{2}$ = $30$ $\frac{m}{s}$

Distance S = $\frac{v^2-u^2}{2a}$ = $\frac{30^2-0^2}{2\times2}$ = $225$ m from starting point

When it accelerates the distance travelled is 900 m. Then again declarates and attain 30 $\frac{m}{s}.$

$\therefore$ u = $60$ $\frac{m}{s}$ , v = $30$ $\frac{m}{s}$ , a = -1 $\frac{m}{s^2}$

Distance = $\frac{v^2-u^2}{2a}$ = $\frac{30^2-60^2}{2(-1)}$ = $1350$ m $\\$ Position is $900$ + $1350$ = $2250$ = $2.25$ $km$ from starting point.

**16.** A bullet travelling with a velocity of $16$ $m/s$ penetrates
a tree trunk and comes to rest in $0.4$ $m$. Find the time
taken during the retardation.

16 None

Solutions$u = 16 \frac{m}{s}$ (initial) , $v =0,$ $s =0.4m$

Deceleration $a$ = $\frac{v^2-u^2}{2s}$ = $-320$ $\frac{m}{s^2}$.

Time = $t$ = $\frac{v-u}{a}$ = $\frac{0-16}{-320}$ = $0.05$ $sec$.

**17.** A bullet going with speed $350$ $m/s$ enters a concrete wall
and penetrates a distance of $5.0$ $cm$ before coming to
rest. Find the deceleration.

17 None

Solutions$u$ = $350$ $\frac{m}{s}$ , $s$ = $5$ $cm$ = $0.05m$ , $v=0$

Deceleration = $a$ = $\frac{v^2-u^2}{2a}$ = $\frac{0-(350)^2}{2\times0.05}$ = $-12.2 \times 10^5$ $\frac{m}{s^2}$.

Deceleration is $12.2 \times 10^5$ $\frac{m}{s^2}.$

**18.** A particle starting from rest moves with constant
acceleration. If it takes $5.0$ $s$ to reach the speed $18.0$
$km/h$ find $\\$(a) the average velocity during this period,
and $\\$(b) the distance travelled by the particle during this
period.

18 None

Solutions$u = 0$ , $v = 18\frac{km}{hr}$ = $5$ $\frac{m}{s}$ , $t$ = $5$ $sec$

$a$ = $\frac{v-u}{t}$ = $\frac{5-0}{5}$ = $1$ $\frac{m}{s^2}.$ $\\$ $s$ = $ut + \frac{1}{2}at^2 = 12.5m$

a) Average velocity $V_{avg} = \frac{(12.5)}{5} = 2.5\frac{m}{s}$. $\\$ b) distance travelled is $12.5 m$

**19.** A driver takes $0.20$ $s$ to apply the brakes after he sees
a need for it. This is called the reaction time of the
driver. If he is driving a car at a speed of $54$ $km/h$ and
the brakes cause a deceleration of $6.0$ $m/s^2$, find the
distance travelled by the car after he sees the need to
put the brakes on

19 None

SolutionsIn reaction time the body moves with the speed $54$ $\frac{km}{hr}$ = $15$ $\frac{m}{sec}$ (constant speed) $\\$ Distance travelled in this time $S_1= 15\times0.2 = 3m$. $\\$ When brakes are applied , $\\$ $u$ = 15 $\frac{m}{s}$, $v$ = 0, $a = -6\frac{m}{s^2}$ (deceleration) $\\$ $S_2 = \frac{v^2-u^2}{2a} = \frac{0-15^2}{2(-6)} = 18.75m$

Total distance $s = s_1 +s_2 = 3+18.75 = 21.75 = 22m$.

**20.** A police jeep is chasing a culprit going on a motorbike.
The motorbike crosses a turning at a speed of $72$ $km/h$.
The jeep follows it at a speed of $90$ $km/h$, crossing the
turning ten seconds later than the bike. Assuming that
they travel at constant speeds, how far from the turning
will the jeep catch up with the bike ?

20 None

Solutions$V_P = 90\frac{km}{h} = 25\frac{m}{s}.$ $\\$ $V_C= 72\frac{km}{h} = 20\frac{m}{s}.$ $\\$ In 10 sec culprit reaches at point B from A. $\\$ Distance converted by culprit $S = vt=20\times10= 200 m$ $\\$ At time $t= 10 sec$ the police jeep is $200m$ behind the culprit. $\\$ Time = $\frac{s}{v} = \frac{200}{5} =40s.$ (Relative velocity is considered).

In $40s$ the police jeep will move from $A$ to a distance $S$, where $S =vt=25\times40=1000m=1.0km $ away. $\\$ $\therefore$ The jeep will catch up the bike , $1km$ far from the turning.

**21.** A car travelling at $60$ $km/h$ overtakes another car
travelling at $42$ $km/h$. Assuming each car to be $5.0$ $m$
long, find the time taken during the overtake and the
total road distance used for the overtake.

21 None

Solutions$V_1 = 60\frac{km}{hr} = 16.6 \frac{m}{s}.$ $\\$ $V_2 = 42\frac{km}{hr} = 11.6 \frac{m}{s}.$ $\\$ Relative velocity between the cars = $(16.6-11.6)= 5\frac{m}{s}.$ $\\$ Distance to be travelled by first car is $5+t=10m.$ $\\$ $Time = t=\frac{s}{v} = \frac{0}{5}=2 sec$ to cross the $2^{nd}$ car.

In $2$ $sec$ the $1^{st}$ car moved = $16.6 \times2=33.2$ $\\$ H also covered its own length $5$ $m$ $\\$ $\therefore$ Total road distance used for the overtake = $33.2+5=38m$

**22.** A ball is projected vertically upward with a speed of
$50$ $m/s$. Find $\\$(a) the maximum height, $\\$(b) the time to
reach the maximum height, $\\$(c) the speed at half the
maximum height. Take g = $10$ $m/s^2$

22 None

Solutions$u$ = $50 \frac{m}{s} , g=-10\frac{m}{s^2}$ When moving upward , $v =0 $ (at highest point)

a) $S= \frac{v^2-u^2}{2a}=\frac{0-50^2}{2(-10)} = 125m$ $\\$ maximum height reached = 125 m

b) $ t=\frac{(v-u)}{a} = \frac{(0-50)}{-10} = 5 sec$ $\\$ c) $s' = \frac{125}{2} = 62.5m$, $u=50\frac{m}{s}, a = -10\frac{m}{s^2}, v^2-u^2=2as$ $\\$

$\Rightarrow$v = $\sqrt{(u^2+2as)} = \sqrt{(50^2+2(-10)(62.5))} = 35\frac{m}{s}$.

**23.** A ball is projected vertically upward with a speed of
$50$ $m/s$. Find $\\$(a) the maximum height, $\\$(b) the time to
reach the maximum height, $\\$(c) the speed at half the
maximum height. Take g = $10$ $m/s^2$

23 None

Solutions**24.** A ball is projected vertically upward with a speed of
$50$ $m/s$. Find $\\$(a) the maximum height, $\\$(b) the time to
reach the maximum height, $\\$(c) the speed at half the
maximum height. Take g = $10$ $m/s^2$

24 None

Solutions**25.** A ball is projected vertically upward with a speed of
$50$ $m/s$. Find $\\$(a) the maximum height, $\\$(b) the time to
reach the maximum height, $\\$(c) the speed at half the
maximum height. Take g = $10$ $m/s^2$

25 None

Solutions**26.** A ball is dropped from a balloon going up at a speed of
$7$ $m/s$. If the balloon was at a height $60$ $m$ at the time
of dropping the ball, how long will the ball take in
reaching the ground ?

26 None

SolutionsInitially the ball is going upward $\\$ $u =-7\frac{m}{s},$ $s=60m$ , $a =g=10\frac{m}{s^2}$ $\\$ $s=ut+\frac{1}{2}at^2$ $\Rightarrow$ $60=-7t+\frac{1}{2}10t^2$

$\Rightarrow$ $5t^2-7t-60=0$

$t$=$\frac{7\pm\sqrt{49-4.5(-60)}}{2\times5}$ = $\frac{7\pm35.34}{10}$

taking positive sign $t$ = $\frac{7+35.34}{10}$ = $4.2 sec$ ( $\therefore$ t $\neq$ -ve)

Therefore , the ball will take 4.2 sec to reach the ground.

**27.** A stone is thrown vertically upward with a speed of
$28$ $m/s$. $\\$(a) Find the maximum height reached by the
stone. $\\$(b) Find its velocity one second before it reaches
the maximum height. $\\$(c) Does the answer of part
(b) change if the initial speed is more than $28$ $m/s$ such
as $40$ $m/s$ or $80$ $m/s$ ?

27 None

Solutions$u =28\frac{m}{s},$ $v=0$ , $a=g=-9.8\frac{m}{s^2}$

a) $S=\frac{v^2-u^2}{2a} =\frac{0-28^2}{2((9.8)} =40m$

b) time $t=\frac{v-u}{a} = \frac{0-28}{-9.8}=2.85$

$t'=2.85-1=1.85$

$v'=u+at'=28-(9.8)(1.85)=9.87\frac{m}{s}$

$\therefore$ The velocity is 9.87 $\frac{m}{s}.$

c) No it will not change. As after one second velocity becomes zero for any initial velocity and deceleration is $g=9.8\frac{m}{s^2}$ remains same. for initial velocity more than $28 \frac{m}{s}$ max height increases.

**28.** A person sitting on the top of a tall building is dropping
balls at regular intervals of one second. Find the
positions of the $3rd$, $4th$ and $5th$ ball when the $6th$ ball
is being dropped.

28 None

SolutionsFor every ball , $u=0 , a=g=9.8\frac{m}{s^2}$ $\\$ $\therefore$ $4^{th}$ ball move for $2$ $sec$, $5^{th}$ ball $1 sec$ and $3^{rd}$ ball $3sec$ when $6^{th}$ ball is being dropped.

For $3^{rd}$ ball $t=3sec$ $\\$

$S_3 = ut+\frac{1}{2}at^2=0+\frac{1}{2}(9.8)3^2=4.9m$ below the top.

For $4^{th}$ ball $t=2sec$ $\\$

$S_2 = 0+\frac{1}{2}gt^2=\frac{1}{2}(9.8)2^2=19.6m$ below the top (u=0).

For $5^{th}$ ball $t=1sec$ $\\$

$S_3 = ut+\frac{1}{2}at^2=0+\frac{1}{2}(9.8)1^2=4.9m$ below the top.

**29.** A healthy youngman standing at a distance of $7$ $m$ from
a $11.8$ $m$ high building sees a kid slipping from the top
floor. With what speed (assumed uniform) should he run
to catch the kid at the arms height $(1.8 m)$ ?

29 None

SolutionsAt point B( i.e. over $1.8m$ from ground) the kid should be catched. $\\$ For kid initial velocity u=0 $\\$ Acceleration = $9.8 \frac{m}{s^2}$ $\\$ Distance $S=11.8-1.8=10m$ $\\$ $S=ut+\frac{1}{2}at^2 \Rightarrow 10=0+\frac{1}{2}(9.8)t^2$ $\\$ $\Rightarrow t^2=2.04 \Rightarrow t=1.42.$

In this time the man has to reach at the bottom of the building. $\\$ Velocity $\frac{s}{t} = \frac{7}{1.42}=4.9\frac{m}{s}.$

**30.** An NCC parade is going at a uniform speed of $6$ $km/h$
through a place under a berry tree on which a bird is
sitting at a height of $12.1$ $m$. At a particular instant the
bird drops a berry. Which cadet (give the distance from
the tree at the instant) will receive the berry on his
uniform ?

30 None

SolutionsLet the time of fall be 't' . initial velocity u=0 $\\$ Acceleration $a= 9.8\frac{m}{s^2}$ Distance $S= \frac{12}{1}m$ $\\$ $\therefore S=ut+\frac{1}{2}at^2$ $\\$ $\Rightarrow 12.1=0+\frac{1}{2}(9.8)\times$ $t^2$

$\Rightarrow$ $t^2=\frac{12.1}{4.9}=2.46 \Rightarrow t=1.57sec$

For cadet velocity = 6 $\frac{km}{hr}=1.66\frac{m}{sec}$ $\\$ Distance = $vt=1.57\times1.66=2.6m$. $\\$ The cadet , $2.6$ m away from tree will receive the berry on his uniform.

$\frac{my \quad name}{name}$

**31.** A ball is dropped from a height. If it takes $0.200$ $s$ to
cross the last $6.00$ m before hitting the ground, find the
height from which it was dropped. Take $g$ = $10$ $m/s^2$.

31 None

SolutionsFor last $6\ m\ distance\ travelled\ s = 6m$

For last $6\ m\ distance\ travelled\ s = 6m\ , u=?$ $\\$ $t=0.2\ sec,\ a=g=9.8\ \frac{m}{s^2}$

$S=ut+\frac{1}{2}at^2 \Rightarrow 6=u(0.2)+4.9\times0.04$ $\\$ $\Rightarrow u=\frac{5.8}{0.2} =29\ \frac{m}{s}.$ $\\$ $For\ distance\ x\ ,\ u=0,\ v=29\ \frac{m}{s},\ a =g=9.8\ \frac{m}{s^2}$ $\\$

$S=\frac{v^2-u^2}{2a} = \frac{29^2-0^2}{2\times 9.8}=42.05\ m.$

$Total\ distance=42.05+6=48.05=48\ m.$

**32.** A ball is dropped from a height of $5$ $m$ onto a sandy floor
and penetrates the sand up to $10$ $cm$ before coming to
rest. Find the retardation of the ball in sand assuming
it to be uniform

32 None

Solutions$Consider\ the\ motion\ of\ ball\ from\ A\ to\ B. \\ B \rightarrow just\ above\ the\ sand\ (just\ to\ penetrate)\\ u=0,\ a=9.8\ \frac{m}{s^2},\ s=5\ m.\\ S=ut+\frac{1}{2}at^2\\ \Rightarrow 5=0+\frac{1}{2}(9.8)t^2\\ \Rightarrow t^2=\frac{5}{4.9}=1.02\Rightarrow t=1.01.\\ \therefore velocity\ at\ B,\ v=u+at=9.8\times 1.01\ (u=0)\ =9.89\ \frac{m}{s}.$

$From\ motion\ of\ ball\ in\ sand$ $\\$ $u_1=9.89\ \frac{m}{s},\ v_1=0,\ $ $\\$ $a=?,\ s=10\ cm=0.1m.$ $\\$ $a=\frac{v_{1}^2-u_{1}^2}{2s}=\frac{0-(9.89)^2}{2\times 0.1}= -490\ \frac{m}{s^2}\\ The\ retardation\ in\ sand\ is\ 490\ \frac{m}{s^2}.$

**33.** An elevator is descending with uniform acceleration. To
measure the acceleration, a person in the elevator drops
a coin at the moment the elevator starts. The coin is $6$
$ft$ above the floor of the elevator at the time it is dropped.
The person observes that the coin strikes the floor in $1$
$second$. Calculate from these data the acceleration of the
elevator

33 None

Solutions$For\ elevator\ and\ coin\ u=0$ $\\$ $As\ the\ elevator\ descends\ downward\ with\ acceleration\ a'\ (say)$ $\\$ $The\ coin\ has\ to\ move\ more\ distance\ than\ 1.8m\ to\ strike\ the\ floor.\ Time\ taken\ t=1\ sec.$ $\\$ $S_c=ut+\frac{1}{2}a't^2=0+\frac{1}{2}g(1)^2=\frac{1}{2}g$ $\\$ $S_e=ut+\frac{1}{2}at^2=u+\frac{1}{2}a(1)^2=\frac{1}{2}a$ $\\$ $Total\ distance\ covered\ by\ coin\ is\ given\ by\ =1.8+\frac{1}{2}a=\frac{1}{2}g$ $\\$ $\Rightarrow1.8+\frac{a}{2}=\frac{9.8}{2}=4.9\\ \Rightarrow=6.2\frac{m}{s^2}=6.2\times3.28=20.34\ \frac{ft}{s^2}.$

**34.** A ball is thrown horizontally from a point $100$ $m$ above
the ground with a speed of $20$ $m/s$. Find $\\$(a) the time it
takes to reach the ground, $\\$(b) the horizontal distance it
travels before reaching the ground, $\\$(c) the velocity
(direction and magnitude) with which it strikes the
ground.

34 None

Solutions$It\ is\ a\ case\ of\ projectile\ fired\ horizontally\ from\ a\ height$ $\\$ $h=100\ m,\ g=9.8\ \frac{m}{s^2}$ $\\$ $a)\ Time\ taken\ to\ reach\ the\ ground\ t\ =\sqrt{(\frac{2h}{g})}$

$=\sqrt{\frac{2\times100}{9.8}}=4.51\ sec.$ $\\$ $b)\ Horizontal\ range\ x=ut = 20\times 4.5=90\ m.$ $\\$ $c)\ Horizontal\ velocity\ remains\ constant\ through\ out\ the\ motion$ $\\$ $At\ A,\ V=20\ \frac{m}{s}\\ AV_y=u+at=0+9.8\times4.5=44.1\ \frac{m}{s}.\\ Resultant\ velocity\ V_r=\sqrt{(44.1)^2+20^2}=48.42\ \frac{m}{s}.$ $\\$ $tan\beta=\frac{ V_y}{V_x}=\frac{44.1}{20}=2.205$

$\Rightarrow \beta=tan^{-1}(2.205) = 60^{\circ}$

$The\ ball\ strikes\ the\ ground\ with\ a\ velocity\ 48.42\ \frac{m}{s}\ at\ an\ angle\ 66^{\circ}\ with\ horizontal.$

**35.** A ball is thrown at a speed of $40$ $m/s$ at an angle of $60°$
with the horizontal. Find $\\$(a) the maximum height
reached and $\\$(b) the range of the ball. $\\$Take $g$ = $10$ $m/s^2$

35 None

Solutions$u=40\ \frac{m}{s},\ a=g=9.8\ \frac{m}{s^2},\ \theta=60^{\circ}\ Angle\ of\ projection.\\ a)\ Maximum\ height\ h=\frac{u^{2}sin^{2}\theta}{2g}=\frac{40^{2}(sin60^\circ)^{2}}{2\times10}=60\ m$

$b)\ Horizontal\ range\ X=\frac{(u^2sin2\theta)}{g}=\frac{(40^2sin2(60^\circ))}{10}=80\sqrt{3}m$

**36.** In a soccer practice session the football is kept at the
centre of the field $40$ $yards$ from the $10$ $ft$ high goalposts.
A goal is attempted by kicking the football at a speed
of $64$ $ft/s$ at an angle of $45°$ to the horizontal. Will the
ball reach the goal post ?

36 None

Solutions$g=9.8\ \frac{m}{s^2},\ 32.2\ \frac{ft}{s^2}\ ;\ 40\ yd=120\ ft\\ horizontal\ range\ x=120\ ft,\ u=64\ \frac{ft}{s},\ \theta = 45^\circ\\ We\ know\ that\ horizontal\ range\ X = u\ cos{\theta}t$

$\Rightarrow=t=\frac{x}{ucos\theta}=\frac{120}{64cos45^\circ}=2.65\ sec$

$y=$u$sin\theta(t)-\frac{1}{2}gt^2=64\frac{1}{\sqrt{2}}(2.65)-\frac{1}{2}(32.2)(2.65)^2$

$=7.08\ ft\ which\ is\ less\ than\ the\ height\ of\ goal\ post.$

In time $2.65,$ the ball travels horizontal distance $120\ ft\ (40\ yd)$ and vertical height $7.08\ ft$ which is less 10 ft. The ball will reach the goal post.

**37.** A popular game in Indian villages is goli which is played
with small glass balls called golis. The goli of one player
is situated at a distance of $2.0$ $m$ from the goli of the
second player. This second player has to project his goli
by keeping the thumb of the left hand at the place of
his goli, holding the goli between his two middle fingers
and making the throw. If the projected goli hits the goli
of the first player, the second player wins. If the height
from which the goli is projected is $19.6\ cm$ from the
ground and the goli is to be projected horizontally, with
what speed should it be projected so that it directly hits
the stationary goli without falling on the ground earlier ?

37 None

Solutions$The\ goli\ moves\ like\ a\ projectile\\ Here\ h=0.196\ m.\\ Horizontal\ distance\ X=2\ m.\\ Acceleration\ g=9.8\ \frac{m}{s^2}.\\ Time to reach the ground i.e. \\ t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times0.196}{9.8}}=0.2\ sec.\\ Horizontal\ velocity\ with\ which\ it\ is\ projected\ be\ u.\\ \therefore x=ut\\ \Rightarrow\ u=\frac{x}{t}=\frac{2}{0.2}=10\ \frac{m}{s}.$

**38.** Figure (3-E8) shows a $11.7$ $ft$ wide ditch with the
approach roads at an angle of $15°$ with the horizontal.
With what minimum speed should a motorbike be
moving on the road so that it safely crosses the ditch ?
Assume that the length of the bike is $5$ $ft$, and it leaves
the road when the front part runs out of the approach
road.

38 None

Solutions$Horizontal\ range\ X=11.7+5=16.7\ ft\ covered\ by\ te\ bike.\\ g=9.8\ \frac{m}{s^2}=32.2\ \frac{ft}{s^2}.$ $\\$ $y=xtan\theta-\frac{gx^2sec^2\theta}{2u^2} \Rightarrow\ u^2=\frac{gx^2sec^2\theta}{2xtan\theta}=\frac{gx}{2sin\theta (cos\theta)}=\frac{gx}{sin2\theta}$ $\\$ $\Rightarrow u=\sqrt{\frac{(32.2)(16.7)}{\frac{1}{2}}}\ (beacuse\ sin30^\circ=\frac{1}{2})$ $\\$ $\Rightarrow u =32.79\ \frac{ft}{s}= 32\ \frac{ft}{s}.$

**39.** A person standing on the top of a cliff $171$ $ft$ high has
to throw a packet to his friend standing on the ground
$228$ $ft$ horizontally away. If he throws the packet directly
aiming at the friend with a speed of $15.0$ $ft/s$, how short
will the packet fall ?

39 None

Solutions$tan\theta=\frac{171}{228}\Rightarrow \theta=tan^{-1}(\frac{171}{228})\\ The\ motion\ of\ projectile\ (i.e.\ the\ packed)\ is\ from\ A.\ Taken\ references\ axis\ at\ A.\\ \therefore\theta=-37^\circ\ as\ u\ is\ below\ x-axis.\\ u=15\ \frac{ft}{s},\ g=32.2\ ft/s^2,\ y=-171\ ft\\ y=xtan\theta-\frac{x^2gsec^2\theta}{2u^2}\\ \therefore -171=-x(0.7536)-\frac{x^2g(1.568)}{2(225)}\\ \Rightarrow 0.1125x^2+0.7536x-171=0\\ x=35.78\ ft\ (can\ be\ calculated)\\ Horizontal\ range\ covered\ by\ the\ packet\ is\ 35.78\ ft.\\ So,\ the\ packet\ will\ fall\ 228-35.78=192\ ft\ short\ of\ his\ friend.$

**40.** A ball is projected from a point on the floor with a speed
of $15$ $m/s$ at an angle of $60°$ with the horizontal. Will it
hit a vertical wall $5$ $m$ away from the point of projection
and perpendicular to the plane of projection without
hitting the floor ? Will the answer differ if the wall is
$22$ $m$ away ?

40 None

SolutionsHere $u=15\ \frac{m}{s},\ \theta=60^\circ,\ g=9.8\ \frac{m}{s^2}\\ Horizontal\ range\ X=\frac{u^2sin2\theta}{g}=\frac{15^2sin(2\times60^\circ)}{9.8}=19.88\ m.$

In first case the wall is $5\ m$ away from projection point, so it is in the horizontal range of projectile . So the ball will hit the wall. In second case (22 m away) wall is not within the horizontal range. So the ball would not hit the wall.

**41.** Find the average velocity of a projectile between the
instants it crosses half the maximum height. It is
projected with a speed $u$ at an angle $0$ with the
horizontal.

41 None

SolutionsTotal of flight T = $\frac{2usin\theta}{g}\\$ Average velocity = $\frac{change\ in\ displacement}{time}$ $\\$ from the figure , it can be said AB is horizontal . So there is no effect of vertical component of the velocity during this displacement. So because the body moves at a constant speed of $u'cos\theta ' $ in horizontal direction. The average velocity during this displacement will be $ucos\theta$ in the horizontal direction.

**42.** A bomb is dropped from a plane flying horizontally with
uniform speed. Show that the bomb will explode
vertically below the plane. Is the statement true if the
plane flies with uniform speed but not horizontally ?

42 None

SolutionsDuring the motion of bomb its horizontal velocity $u$ remains constant and is same as that of aeroplane at every point of its path. Suppose the bomb explode i.e. reach the ground in time t. Distance travelled in horizontal direction by bomb $= ut = $ the distance travelled by aeroplane. so bomb explode vertically below the aeroplane. Suppose the aeroplane move making angle $\theta $ with horizontal. For both bomb and aeroplane , horizontal distance is u$cos\theta$t . t is time for bomb to reach the ground. so in this case also , the bomb will explode vertically below aeroplane.

**43.** A boy standing on a long railroad car throws a ball
straight upwards. The car is moving on the horizontal
road with an acceleration of $1$ $m/s^2$ and the projection
velocity in the vertical direction is $9.8$ $m/s$. How far
behind the boy will the ball fall on the car ?

43 None

SolutionsLet the velocity of car be u when the all is thrown. Initial velocity of car is = Horizontal velocity of ball. Distance travelled by ball B $S_b=ut$ (in horizontal direction) And by car $S_c = ut +\frac{1}{2}at^2$ where $t\rightarrow time\ of\ flight\ of\ ball\ in\ air.\\ \therefore$ Car has travelled extra distance $S_c-S_b=\frac{1}{2}at^2.\\ $ Ball can be considered as a projectile having $\theta = 90^\circ$. $\therefore t=\frac{2usin\theta}{g}=\frac{2\times9.8}{9.8}=2\ sec.\\ \therefore S_c-S_b=\frac{1}{2}at^2=2\ m.\\$ $\therefore$ The ball will drop $2m$ behind the body.

**44.** A staircase contains three steps each $10$ $cm$ high and
$20$ $cm$ wide (figure 3-E9). What should be the minimum
horizontal velocity of a ball rolling off the uppermost
plane so as to hit directly the lowest plane ?

44 None

SolutionsAt minimum velocity it will move just touching point E reaching the ground. $A$ is origin of reference coordinate. it $u$ is the minimum speed. $X=40,\ Y=-20,\ \theta=0^\circ\\ \therefore Y=xtan\theta-g\frac{x^2sec^2\theta}{2u^2}\ (because\ g=10\ \frac{m}{s^2}=1000\ \frac{cm}{s^2})\\ \Rightarrow -20=xtan\theta-\frac{1000\times40^2\times1}{2u^2}\\ \Rightarrow u=200\ \frac{cm}{s}=2\ m/s\\ \therefore The minimum horizontal velocity is 2\ m/s.$

**45.** A person is standing on a truck moving with a constant
velocity of $14.7$ $m/s$ on a horizontal road. The man throws
a ball in such a way that it returns to the truck after
the truck has moved $58.8$ $m$. Find the speed and the
angle of projection $\\$(a) as seen from the truck, $\\$(b) as seen
from the road

45 None

Solutionsa) As seen from the truck the ball moves vertically upwards comes back. Time taken = time taken by truck to cover $58.8\ m$. $\therefore time= \frac{s}{v}=\frac{58.8}{14.7}=4\ sec.\ (V=14.7\ m/s\ of\ truck)\\ u=?,\ v=0,\ g=-9.8\ m/s^2\ (goind\ upward),\ t=\frac{4}{2}=2\ sec.\\ v=u+at\Rightarrow 0=u-9.8\times2\Rightarrow u=19.6\ m/s./ (vertical\ upward\ velocity).$

b) From road it seems to be projectile motion . Total time of flight = $4\ sec$ In this time horizontal range covered $58.8\ m=x \\ \therefore X=ucos\theta t\\ \Rightarrow ucos\theta=14.7 ..(1)\\$

Taking vertical component of velocity into consideration. $\\$ $y=\frac{0^2-(19.6)^2}{2\times(-9.8)}= 19.6\ m\ [from\ (a)]\\ \therefore y=usin\theta t-\frac{1}{2}gt^2\\ \Rightarrow 19.6=usin\theta(2)-\frac{1}{2}(9.8)2^2\Rightarrow2usin\theta=19.6\times2\\ \Rightarrow usin\theta=19.6 ...(ii)\\ \frac{usin\theta}{ucos\theta}=tan\theta\Rightarrow\frac{19.6}{14.7}=1.333\\ \Rightarrow \theta=tan^{-1}(1.333)=53^\circ\\$ Again $ucos\theta=14.7\\ \Rightarrow u=\frac{14.7}{ucos53^\circ}=24.42\ m/s.\\$ The speed of ball is $24.42\ m/s$ at an angle $53^\circ$ with horizontal as seen from the road.

**46.** The benches of a gallery in a cricket stadium are $1\ m$
wide and $1\ m$ high. A batsman strikes the ball at a level
one metre above the ground and hits a mammoth sixer.
The ball starts at $35\ m/s$ at an angle of $53°$ with the
horizontal. The benches are perpendicular to the plane
of motion and the first bench is $110\ m$ from the batsman.
On which bench will the ball hit ?

46 None

Solutions$\theta=53^\circ,\ so cos53^\circ = \frac{3}{5}\\$ $Sec^{2}\theta=\frac{25}{9}\ and\ tan\theta=\frac{4}{3}\\$ Suppose the ball lands on nth bench $\\$ So, $y=(n-1)1\ ...(1)\ $[ball starting point $1\ m$ above ground] $\\$ Again $y=xtan\theta-\frac{gx^2sec^2\theta}{2u^2}\ [x=110+n-1=110+y] \\ \Rightarrow y=(110+y)(\frac{4}{3})-\frac{10(110+y)^2(\frac{25}{9})}{2\times35^2}\\ \Rightarrow=\frac{440}{3}+\frac{4}{3}y-\frac{250(110+y)^2}{18\times35^2}\\$

From the equation , y can be calculated.$\\$ $\therefore y =5\\ \Rightarrow n-1=5\Rightarrow n=6.\\$ The ball will drop in sixth bench.

**47.** A man is sitting on the shore of a river. He is in the
line of a $1.0$ $m$ long boat and is $5.5$ $m$ away from the
centre of the boat. He wishes to throw an apple into the
boat. If he can throw the apple only with a speed of $10$
$m/s$, find the minimum and maximum angles of
projection for successful shot. Assume that the point of projection and the edge of the boat are in the same
horizontal level.

47 None

SolutionsWhen the apple just touches the end B of the boat. $\\$ $ x=5\ m,\ u=10\ m/s,\ g=10\ m/s^2,\ \theta=?\\ x=\frac{u^2sin2\theta}{g}\\ \Rightarrow 5=\frac{10^2sin2\theta}{10}\Rightarrow 5=10\ sin\ 2\theta\\ \Rightarrow sin2\theta=\frac{1}{2}\Rightarrow sin\ 30^\circ\ or\ sin150^\circ\\ \Rightarrow \theta=15^\circ\ or\ 75^\circ\\$ Similarly for end C, $x=6\ m$ $\\$ $Then\ 2\theta_1=sin^{-1}(\frac{gx}{u^2})=sin^{-1}(0.6)=182^\circ\ or\ 71^\circ .\\$ So, for a successful shot, $\theta$ may vary from $15^\circ\ to\ 18^\circ\ or\ 71^\circ\ to\ 75^\circ\ .$

**48.** A river $400$ $m$ wide is flowing at a rate of $2.0$ $m/s$. A
boat is sailing at a velocity of $10$ $m/s$ with respect to the
water, in a direction perpendicular to the river. $\\$(a) Find
the time taken by the boat to reach the opposite bank.$\\$
(b) How far from the point directly opposite to the
starting point does the boat reach the opposite bank ?

48 None

Solutionsa) Here the boat moves with the resultant velocity R. But the vertical component $10\ m/s$ takes him to the opposite shore. $\\$ $tan\ \theta=\frac{2}{10}=\frac{1}{5}\\ Velocity\ = 10\ m/s\\ distance=400\ m\\ Time=\frac{400}{10}=40\ sec.\\$

b) The boat will reach at point C. $\\ In\ \Delta ABC,\ tan\ \theta=\frac{BC}{AB}=\frac{BC}{400}=\frac{1}{5}\\ \Rightarrow BC=\frac{400}{5}=80\ m.$

**49.** A swimmer wishes to cross a $500$ $m$ wide river flowing
at $5$ $km/h$. His speed with respect to water is $3\ km/h.$ $\\$
(a) If he heads in a direction making an angle 0 with
the flow, find the time he takes to cross the river.
$\\$(b) Find the shortest possible time to cross the river

49 None

Solutionsa) The vertical component $3sin\theta $ takes him to opposite side. $\\$ $Distance =0.5\ km,\ Velocity =3sin\theta\ km/h\\ Time=\frac{Distance}{Velocity}=\frac{0.5}{3sin\theta}hr\\ =\frac{10}{sin\theta}\ min.$

b) Here vertical component of velocity i.e. $3\ km/hr$ takes him to opposite side. $\\$ $TIme=\frac{Distance}{Velocity}=\frac{0.5}{3}=0.16\ hr\\ \therefore 0.16\ hr=60\times0.16=9.6=10\ minute.$

**50.** Consider the situation of the previous problem. The man
has to reach the other shore at the point directly
opposite to his starting point. If he reaches the other
shore somewhere else, he has to walk down to this point.
Find the minimum distance that he has to walk.

50 None

SolutionsVelocity of man $\vec V_{m}=3\ km/hr$ $\\$ BD horizontal distance for resultant velocity R.$\\$ X-component of resultant $R_x=5+3\ cos\theta\\ t=\frac{0.5}{3sin\theta}\\$ which is same for horizontal component of velocity. $\\$ $H=BD=(5+3cos\theta)(\frac{0.5}{3\ sin\theta})=\frac{5+3cos\theta}{6sin\theta}\\$ $For\ H\ to\ be\ min\ (\frac{dH}{d\theta})=0\\ \Rightarrow \frac{d}{d\theta}(\frac{5+3cos\theta}{6sin\theta})=0\\$ $\Rightarrow -18(sin^2\theta+cos^2\theta)-30\ cos\theta=0\\$ $\Rightarrow -30cos\theta=18 \Rightarrow cos\theta=-\frac{18}{30}=\ -\frac{3}{5}\\ Sin\ theta=\sqrt{1-cos^2\theta}=\frac{4}{5}\\$ $\therefore H=\frac{5+3cos\theta}{6sin\theta}=\frac{5+3(-\frac{3}{5})}{6\times(\frac{4}{5})}=\frac{2}{3}\ km.$

**51.** An aeroplane has to go from a point $A$ to another point
$B,\ 500\ km$ away due $30°$ east of north. A wind is blowing
due north at a speed of $20\ m/s$. The air-speed of the
plane is $150\ m/s$. $\\$(a) Find the direction in which the pilot should head the plane to reach the point B. $\\$(b) Find
the time taken by the plane to go from $A$ to $B$.

51 None

SolutionsIn resultant direction $\vec R$ the plane rech the point $B$.$\\$ $Velocity\ of\ wind\ \vec V_w=20\ m/s\\$ $Velocity\ of\ aeroplane\ \vec V_a=150\ m/s\\ In\ \Delta ACD\ according\ to\ sine\ formula$

$\therefore \frac{20}{sin\ A}= \frac{150}{sin\ 30^\circ} \Rightarrow sinA= \frac{20}{150}sin30^\circ= \frac{20}{150}\times \frac{1}{2}= \frac{1}{15}\\ \Rightarrow A=sin^{-1}(\frac{1}{15})$

a) The direction is $sin^{-1}(\frac{1}{15})$ east of the line $AB$ $\\$

b) $sin^{-1}(\frac{1}{15})=3^\circ 48'\\ \Rightarrow 30^\circ+ 3^\circ 48'=33^\circ 48'\\ R=\sqrt{150^2+20^2+2(150)20cos33^\circ 48'}=167\ m/s.\\ Time =\frac{s}{v}=\frac{500000}{167}=2994\ sec=49=50\ min.$