Question

A balloon moves up vertically such that if a stone is projected with a horizontal velocity u relative to balloon, the stone always hits the ground at a fixed point at a distance $\frac{2u^2}{g}$ horizontally away from it. Find the height of the balloon as a function of time.

Answer 1

Time of flight of stone to reach ground:
$t=\sqrt{\frac{2h}{g}}$
where h = height of the balloon
Horizontal range of the stone:
$R=ut$
$\frac{2u^2}{g}=u\sqrt{\frac{2h}{g}}$
$\frac{2u}{g}=\sqrt{\frac{2h}{g}}$
$\frac{4u^2}{g^2}=\frac{2h}{g}$
$h=\frac{2u^2}{g}$
But, $\frac{2u^2}{g}=R=ut$
So,
$h=ut$