Question

A man in a balloon rising vertically with an acceleration of $5m/s^2$ releases a ball $2$ seconds after the balloon is let go from the ground. The greatest height (in $m$) above the ground reached by the ball is: ($g=10m/s^2$)

Answer 1

Formula used: $s=ut+\frac{1}{2}g{t}^2$, $v=u+at$
Height ascended by balloon in $2s$:
$s_1=\frac{1}{2}\times 5\times (2{)}^2=10m$
Velocity of balloon after $2s$:
$v_1=5\times 2=10m/s$
This will also be the velocity of ball when released from balloon.
So further maximum height ascended by ball
$s_2=\frac{(10{)}^2}{(2\times 10)}=5m$
Total height:
$s=s_1+s_2=10+5=15m$

Final answer: 15 m.