Question

A man in a balloon rising vertically with an acceleration of $4.9{\text{m/s}}^2$ releases a ball $2\text{s}$ after the balloon has left the ground. The greatest height above the ground, reached by ball is $(g=9.8{\text{m/s}}^2)$

• A. $14.7\text{m}$
• B. $19.6\text{m}$
• C. $9.8\text{m}$
• D. $24.5\text{m}$

Answer 1

The correct option is A $14.7\text{m}$

$a_m=4.9m/s^2$
$t=2sg=9.8m/s^2$


The ball will have certain upward velocity when released from balloon, let that be $u_b$. So the ball is now in motion under gravity, it will reach a certain height ($h$) and fall. So the total height reached by ball is $S=H+h$.

Here, $v_m=u_b$.
Using, $v_m=u_m+at$
$v_m=0+4.9\times 2=9.8m/s=u_b$
Also, $H=\frac{1}{2}a{t}^2=\frac{1}{2}\times 4.9\times 2^2=9.8m$

Now, for the final velocity of the ball is zero: $v_b=0$
Using $v^2=u^2-2gh$
$0={9.8}^2-2\times 9.8\times h$
$\Rightarrow h=4.9m$

Total height reached by ball :$S=H+h=9.8+4.9=14.7m$