Question

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

• A. 5 m/s
• B. 12 m/s
• C. 3 m/s
• D. 4 m/s

Answer 1

The Correct Option is A

To solve this problem, we need to apply the equations of motion under gravity. Here's a step-by-step explanation:

1. Firstly, we consider the body dropped from a height \( h \) with zero initial velocity.
2. The final velocity \( v_1 \) with which it strikes the ground can be found using the equation of motion: v^2 = u^2 + 2gh. Here, \( u = 0 \), \( g \) is the acceleration due to gravity, and \( v = 3 \, \text{m/s} \) as given.
3. So, substituting the given data, we have: 3^2 = 0 + 2gh, which simplifies to 9 = 2gh.
4. Now, consider the second body that is dropped from the same height \( h \) with an initial velocity \( u_2 = 4 \, \text{m/s}\).
5. To find the final velocity \( v_2 \) of the second body, we again use the equation of motion: v^2 = u^2 + 2gh.
6. Substituting for the second body, we have: v_2^2 = 4^2 + 2gh = 16 + 9 (since \( 2gh = 9 \) from earlier calculation).
7. This gives us v_2^2 = 25, and therefore v_2 = \sqrt{25} = 5 \, \text{m/s}.

Thus, the final velocity of the second mass with which it strikes the ground is 5 m/s. Therefore, the correct answer is 5 m/s.