Question

A stone is dropped from a height of $300$m and at the same time another stone is projected vertically upwards with a velocity of $100$ $m{s}^{-1}$. Find when and where the two stones meet.

Answer 1

height covered by the projectile from the ground

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

$h=100t-\frac{1}{2}\times 9.8\times t^2$

$h=100t–4.9t^2\dots \left(1\right)$

Distance covered by the projectile thrown down in time t will be, since it falls down from the rest$u=0$

$300–h=\frac{1}{2}g{t}^2$

$300–(100t–4.9t^2)=4.9t^2\dots byequation\left(1\right)$

$300=100t,t=3s$

Thus particles meet each other at $3s$
Height at which particles meet

$h=100t–4.9t^2$

$h=100\times 3–4.9\times 9$

$h=255.9m$