Question

A body is dropped from a balloon which is moving upwards with a speed of 10 m/s. The body is dropped when the balloon is at a height of 390 m from the surface of the earth. Find the approximate time taken by the body to reach the surface of the earth. Take $g=10m/s^2$

Answer 1

$u=-10m/s$
[$\because$ The balloon is moving up and the body is dropped down]
$g=10m/s^{2}h=390m$
Using, $h=ut+\frac{1}{2}g{t}^2$
$390=(-10)t+\frac{1}{\text{/}2}\times \text{/}{10}^5\times t^{2}390=5t^2-10t\Rightarrow t^2-2t-78=0\therefore t=\frac{-(-2)\pm \sqrt{(-2{)}^2-4(1\left)\right(-78)}}{2\times 1}=\frac{2\pm \sqrt{4+312}}{2}=\frac{2\pm 17.776}{2}$
We take $t=\frac{2+17.776}{2}$
$=9.888t\approx 10s$
$\therefore$ The body will take approximately 10 s to reach the ground.