Question

Ramesh throws a ball from the top of an 80 m tall tower. The ball falls on the floor and bounces to a height of 8 m. When this ball touches the ground to bounce, exactly at that time a second ball is thrown from the top. Consider g = 10 ms¹.

i) How much time is taken by the 1st ball to touch the ground and bounce to a height of 8 m? ii) When the 1st ball falls and bounces to a maximum height of 5 m, what is the

location of the 2nd ball thrown from above?

Answer 1

80 = (1/2)g t^2 = 5 t^2
so t = 4 seconds for first ball to hit ground

with what speed does it hit?
v = 10 t = 10 * 4 = 40 m/s
assuming no energy lost in collision with floor:
Vi = initial speed up = 40 m/s
h = 8 m
v = Vi - 10 t
h = 40 t - 5 t^2
8 = 40 t - 5 t^2
5 t^2 - 40 t + 8 = 0
t up = 0.205 seconds
total first ball time to reach 8 meters = 4.205 seconds
for 8, not 5 , meters, drop second ball for 0.205 seconds
h = 80 - 5 t^2 = 80 - 5 (0.205)^2

If they mean 5 meters is the max the first ball will reach, then it lost a lot of energy in the crash
how fast did it leave the floor?
v = Vi - 10 t
at top v = 0
Vi = 10 t
h = Vi t - 5 t^2
5 = Vi t- 5 t^2
5 = 10 t^2 - 5 t^2 = 5 t^2
t =1 second upward
so the second one falls for 1 second
h = 80 - 0 t - 5 t^2 = 80 - 5 = 75 meters