Question

A man in a balloon rising vertically with an acceleration of $4.9m/s^2$ releases a ball $2s$ after the balloon is let go from the ground. The greatest height above the ground reached by the ball is $(g=9.8m/s^2)$.

• A. $14.7m$
• B. $19.6m$
• C. $9.8m$
• D. $24.5m$

Answer 1

The correct option is A $14.7m$

By using first equation of motion, velocity of balloon after $2s$ will be
$v=u+at$
$\Rightarrow v=4.9\times 2(\because u=0)$
$\Rightarrow v=9.8m/s$
Height of the balloon at $t=2s$ is given by
$s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s=\frac{1}{2}\times 4.9\times 4$
$=9.8m$
At the time of release, velocity of ball will same as that of balloon
$\Rightarrow v_{\text{ball}}=9.8m/s$ (Upwards)
Further height achieved by the ball after the release is given by
$v^2-u^2=2gh$
$\Rightarrow \frac{u^2}{2g}=h$ ($\because$ final velocity is 0 at maximum height)
$\therefore h=4.9m$
$\therefore$ Maximum height of the ball from the ground $=h_{\text{max}}=9.8+4.9=14.7m$