Question

A stone is dropped from a height of 100m on earth. At the same time, another stone is thrown vertically upwards from a ground with a velocity of 50$\frac{m}{sec}$. At what height from the ground will the stone meet?

Answer 1

Let distance measured from bottom be $s_1$ and distance from top be $s_2$.

$s_2$ = $\frac{1}{2}$ g$t^2$

$s_1$ = ut - $\frac{1}{2}$ g$t^2$

100 = $s_1$ + $s_2$

100 = ut = 50 t

t = 2 secs

Distance from top = $\frac{1}{2}$ g$t^2$ = 19.62 m

Height = 100 - 19.62 = 80.38 m.