Question

A man in a ballon rising vertically with an acceleration of $4.9\frac{m}{se{c}^2}$ 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 $(g=\frac{9.8m}{se{c}^2})$

• 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
$v=at=4.9\times 2=9.8\frac{m}{s}$
Now if the ball is released from the balloon then it acquire same velocity 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 = $h_1+h_2=9.8+4.9=14.7m$