Question

A small ball is released from rest from a point that is 40m above horizontal ground. The ball bounces on the ground and rebounds vertically. Each time the ball bounces on the ground, the speed of the ball is instantaneously reduced by 50%. The ball is modelled as a particle moving freely under gravity, from the instant when it is released until it first hits the ground, and between each successive bounce.
(a) Find the time from the instant when the ball is released from rest to the instant when it hits the ground for the second time.
(b) Find the maximum height reached by the ball above the ground after the ball's third bounce. ​

Answer 1

Answer:

(a) 5.714s

(b) 0.625m

Step-by-step explanation:

Acceleration due to gravity=a=±9.8m/s/s (+ when falling, - when going upwards)

(a)

STAGE A: between start and first bounce

Initial speed=u=0m/s, Distance=s=40m, and Final velocity=v

v²=u²+2as

v²=0²+2(10)(40)

v²=794

v=28m/s (As the ball hits the ground)

v=u+at

28=0+9.8t$_{a}$

t$_{a}$=28/9.8=2.85714285714s

STAGE B: from the first bounce until it starts to fall

u=half the speed it hit the ground with=(28/2)=14m/s, and v=0m/s(when it starts to fall)

v=u+at

0=14-9.8t$_{b}$

t$_{b}$=14/9.8=1.42857142857s

STAGE C: between when it starts to fall after bounce 1, and bounce 2

We can assume that the time it takes to go from the ground to max height after bounce 1 is equal to the time it takes to fall from that same height to the ground. Therefore, t $_{c}$=1.42857142857s

The time from when the ball was released to when it hit the ground for the second time = t

t = t$_{a}$ + t$_{b}$ + t$_{c}$

t=2.85714285714+1.42857142857+1.42857142857

t=5.71428571428s≈5.714s

(b)

Before bounce 1, u=28m/s (see stage a)

After bounce 1, u= 28/2= 14m/s

After bounce 2, u= 14/2 = 7m/s

After bounce 3, u= 7/2 =3.5m/s

u=3.5m/s and v=0m/s(when it starts to fall)

v²=u²+2as

0²=3.5²+2(-9.8)s

19.6s=12.25

s=0.625m (max height after third bounce)