Question

A man in a balloon rising vertically with an acceleration of $4.9\, m / \sec ^{2}$, 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

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