Question

A man in a balloon rising vertically with an acceleration of 4.9 m/$s^2$ releases a ball 2 sec 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. 14.7 m
• B. 19.6 m
• C. 9.8 m
• D. 24.5 m

Answer 1

The correct option is A 14.7 m

Height travelled by ball (with balloon) in 2 sec
$h_1=\frac{1}{2}a{t}^2=\frac{1}{2}\times 4.9\times 2^2=9.8m$
Velocity of the balloon after 2 sec
v= a t=4.9$\times$2=9.8 m/s
Now if the ball is reached from the balloon then it acquired same
velocity in upward direction.
Let it move in upward direction.
Let it move up to maximum height $h_2$
$v^2=u^2-2g{h}_2\Rightarrow 0=(9.8{)}^2-2\times (9.8)\times h_2$
$\therefore h_2$=4.9m
Greatest height above the ground reached by the ball
$=h_1+h_2$=9.8+4.9=14.7m