AIIMS2018Physics-Kinematics

AIIMS 2018 Physics Equations of Motion MCQ Question

Type: MCQ-numerical-Medium-Class 11

A ball is dropped from a high rise platform at t=0t = 0 starting from rest. After 6 sec6\text{ sec} another ball is thrown downwards from the same platform with a speed vv. The two balls meet at t=18 st = 18\text{ s}. What is the value of vv? (take g=10 m/s2g = 10\text{ m/s}^2)

A

75 m/s75\text{ m/s}

B

55 m/s55\text{ m/s}

C

40 m/s40\text{ m/s}

D

60 m/s60\text{ m/s}

Correct Answer

Option A

Detailed Explanation

In the problem, both balls travel the same distance, which is confirmed by the calculations showing that the first ball covers 1620 m in 18 seconds under constant acceleration, while the second ball, starting with an initial velocity vv, covers the same distance in 12 seconds with the same acceleration. The derived velocity v=75m/sv = 75 \, \text{m/s} for the second ball is consistent with the distance formula, demonstrating that the initial conditions and time intervals lead to a valid solution. Since all other options are not provided or relevant to the calculations, they cannot be considered correct.

Found an issue with this question?