Final Velocity in Free Fall – Rankers Physics
Topic: Kinematics
Subtopic: Motion Under Gravity

Final Velocity in Free Fall

A body dropped from a height \(h\) with initial velocity zero, strikes the ground with a velocity \(3\text{ m/s}\)). Another body of same mass dropped from the same height \(h\) with an initial velocity of \(4\text{ m/s}\)). The final velocity of second mass, with which it strikes the ground is:

(1996)

\(5\text{ m/s}\)
\(12\text{ m/s}\)
\(3\text{ m/s}\)
\(4\text{ m/s}\)

Solution:

Concept: Equations of motion under constant gravity.
Formula: \(v_f^2 = v_i^2 + 2gh\).
For the first body: \(v_i = 0\), \(v_f = 3\text{ m/s}\). So, \(3^2 = 0^2 + 2gh \Rightarrow 2gh = 9\).
For the second body: \(v_i = 4\text{ m/s}\). The final velocity is \(v_f'\).
\((v_f')^2 = 4^2 + 2gh = 16 + 9 = 25\).
Therefore, \(v_f' = \sqrt{25} = 5\text{ m/s}\).

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