Question

A man in a balloon rising vertically with an acceleration of $4.9m/{\sec }^2$, releases a ball $2$ seconds after the balloon is let go from the ground. The greatest height above the ground reached by the ball is (in $m$ ) ?

Answer 1

Answer: 14.70

Height risen by balloon in $2\sec$,
$s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s=(0)(2)+\frac{1}{2}(4.9)(2{)}^2$
$=9.8m$
velocity of balloon at the end of $2\sec$,
$v=u+at$
$\Rightarrow v=0+(4.9)(2)=9.8m/s$
Due to inertia when the ball is dropped, it will move upwards with initial velocity $9.8m/s$ for some time and then fall down.
So, for ball for subsequent, upward motion,
$u=+9.8,a=-9.8m/s^2,v=0$
$v^2=u^2+2as$
$0^2=(9.8{)}^2+2(-9.8)(s)$
$s=\frac{9.8}{2}=4.9m$
$\therefore$ Maximum height of ball above ground
$=9.8+4.9=14.7m$