Question

A ball which is thrown vertically upwards reaches the roof of a house $100m$ 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?
($g=9.8m/s^2$)

Answer 1

at max height $v\left(f\right)=0m/s$

$a=-g=-9.8m/s^2$ ... [taking UP as the positive direction]

$s=100m$

$v(f{)}^2=v(i{)}^2+2as$

$v(i{)}^2=v(f{)}^2-2as$

$v\left(i\right)=\sqrt{[0-2\times -9.8\times 100]}$

$v\left(i\right)=44.27m/s$

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]

$s=v\left(i\right)t+\left(1/2\right)a{t}^2$

so for the 1st ball:

$s=44.27t-4.9t^2$

and for the second ball:

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

so $s=4.9t^2$

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

height the 2nd ball is above the ground when the two balls meet $=100-4.9t^2$ ... [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 $=100-4.9\times (100/44.27{)}^2=75m$

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