Question

A hot air balloon, released from ground, moves up with a constant acceleration of $5{\text{ms}}^{-2}$. A stone is released from it at the end of $8s$. Find the height (in$\text{m}$) of the stone from the ground, $8s$ after it is released from the balloon. (Take $g=10{\text{ms}}^{-2}$)

Answer 1

The velocity of the stone, when it is released from the balloon is given by, $v=u+at$ and is equal to the velocity of the balloon. Consider downward direction as position and substitute,
$u=0,a=-5{\text{ms}}^{-2},t=8$
$v=-5\times 8=-40{\text{ms}}^{-1}$
The displacement of the stone after $8s$ is
$s=ut+\frac{1}{2}g{t}^2$
Substituting $s=h,u=-40{\text{ms}}^{-1}$
$t=8s,g=10{\text{ms}}^{-2}$
$h=-40\times 8+\frac{1}{2}\times 10\times 8^2$
$h=-320+320=0$

The stone returns back to the position from where it was released, i.e, the height of the balloon from the ground at the end of $8s$.
$s=ut+\frac{1}{2}a{t}^2$
substitute, $u=0$
$a=-5{\text{ms}}^{-2},t=8s$
$s=-\frac{1}{2}\times 5\times 8\times 8=-5\times 4\times 8$
$s=-160\text{m}$
$\therefore$ The height of the balloon from the ground is $160\text{m}$