Question

A ball which is thrown vertically upwards reaches the roof of a house 100 m high. At the moment this ball is thrown vertically upward, another ball is dropped from rest vertically downwards from the roof of the house. At which height will the balls pass each other? (g 9.8 m/s

Answer 1

First, find the initial velocity the first ball had when it was thrown vertically up: 

at max height v(f) = 0 m/s 

a = -g = -9.8 m/s² ... [taking UP as the positive direction] 

s = 100 m 

v(f)² = v(i)² + 2as 

v(i)² = v(f)² - 2as 

v(i) = √[0 - 2 * -9.8 * 100] 

v(i) = 44.27 m/s 


When the two balls meet their distances above the level the first ball was thrown from (assumed to be the ground ... b/c no other info has been given about it) are equal 

The 2nd ball is dropped at the same time the 1st ball is thrown ... so the times taken to the meeting point for the balls are also equal 


For the 2nd ball, the initial vertical velocity is 0 m/s ... [b/c it is just dropped] 

s = v(i)t + (1/2)at² 

so for the 1st ball: 

s = 44.27t - 4.9t² 


and for the second ball: 

-s = 0 - 4.9t² ... [-s b/c the 2nd ball travels in the negative direction] 

so s = 4.9t² 

BUT we want the distance the 2nd ball is above the ground so 

the height the 2nd ball is above the ground when the two balls meet = 100 - 4.9t² ... [Eqn 1] 

then when the two balls meet: 

44.27t - 4.9t² = 100 - 4.9t² 

44.27t = 100 

t = 100/44.27 


subs t = 100/44.27 into [eqn 1]: 

distance above the ground where the two balls meet = 100 - 4.9 * (100/44.27)² = 75 m 

so the balls pass each other at a height of 75m