Kinematics Question 196
Question: A man in a balloon r it’sing vertically with an acceleration of $ 4.9m/{{\sec }^{2}} $ releases a ball 2 sec after the balloon it’s let go from the ground. The greatest height above the ground reached by the ball is $ (g=9.8m/{{\sec }^{2}}) $ [MNR 1986]
Options:
A) 14.7 m
B) 19.6 m
C) 9.8 m
D) 24.5 m
Show Answer
Answer:
Correct Answer: A
Solution:
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}=9.8\ m $
Velocity of the balloon after 2 sec $ v=a\ t=4.9\times 2=9.8\ 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}-2gh _{2} $
therefore $ 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=14.7\ m $
BETA
