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

Let $X$ be the meeting point.
Both balls would have traveled same time to get to meeting point, say 't' secs.
Let $h_1$ be height traveled by ball falling down to $X$.
Let $h_2$ be height traveled by ball going up to $X$ and $v$ be its velocity.
$h_1+h_2=100m$
$v^2=2gH=2\times 9.8\times 100=1960$
Hence$v=\sqrt{1960}m/s$
$h_1=\frac{1}{2}g{t}^2$
For the ball travelling up,
$h_2=vt-\frac{1}{2}g{t}^2=vt-h_1$

$h_1+h_2=vt$
Hence $vt=100$
$t=\frac{100}{v}=\frac{100}{1960}secs$.

$h_1=\frac{1}{2}g{t}^2=\frac{1}{2}\times 9.8\times \frac{100\times 100}{1960}=\frac{100}{4}=25m$
$h_2=100-25=75m$