SATHEE: Kinematics Question 196

Kinematics Question 196

Question: A man in a balloon r it’sing vertically with an acceleration of $4.9 m / \sec^{2}$ 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 $(g = 9.8 m / \sec^{2})$ [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 $h_{2}$

$v^{2} = u^{2} - 2 g h_{2}$

therefore $0 = {(9.8)}^{2} - 2 \times (9.8) \times h_{2}$
$∴$ $h_{2}$ =4.9m

Greatest height above the ground reached by the ball $= h_{1} + h_{2} = 9.8 + 4.9 = 14.7 m$

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Kinematics Question 196

Question: A man in a balloon r it’sing vertically with an acceleration of 4.9m/sec2 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/sec2) [MNR 1986]

Options:

Answer:

Solution:

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Question: A man in a balloon r it’sing vertically with an acceleration of $4.9 m / \sec^{2}$ 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 $(g = 9.8 m / \sec^{2})$ [MNR 1986]