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 is thrown from the same height h with an initial velocity of 4 m/s. Find the final velocity of second mass, with which it strikes the ground.

Options

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

Answer

To solve the problem, we will use the equations of motion under uniform acceleration due to gravity.

### Step-by-Step Solution:

1. **Identify the first case**: A body is dropped from height \( h \) with an initial velocity \( u_1 = 0 \) m/s and strikes the ground with a final velocity \( v_1 = 3 \) m/s.

2. **Use the equation of motion**: We can use the equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration (which is \( g \), the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \))
- \( s \) = distance (which is \( h \))

For the first case:
\[ (3)^2 = (0)^2 + 2gh \]
This simplifies to:
\[ 9 = 2gh \quad (1) \]

3. **Identify the second case**: A second body of the same mass is thrown from the same height \( h \) with an initial velocity \( u_2 = 4 \) m/s. We need to find its final velocity \( v_2 \) when it strikes the ground.

4. **Use the same equation of motion for the second case**:
\[ v_2^2 = u_2^2 + 2gh \]
Substituting \( u_2 = 4 \) m/s:
\[ v_2^2 = (4)^2 + 2gh \]
This simplifies to:
\[ v_2^2 = 16 + 2gh \quad (2) \]

5. **Substitute \( 2gh \) from equation (1)** into equation (2):
From equation (1), we know:
\[ 2gh = 9 \]
Therefore, substituting this into equation (2):
\[ v_2^2 = 16 + 9 \]
This simplifies to:
\[ v_2^2 = 25 \]

6. **Calculate \( v_2 \)**:
\[ v_2 = \sqrt{25} = 5 \, \text{m/s} \]

### Final Answer:
The final velocity of the second mass when it strikes the ground is \( 5 \, \text{m/s} \).

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