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 4m/s. Find the final velocity of second mass with which it strikes the ground.

Options

A. `3ms^(-1)`
B. `4ms^(-1)`
C. `5ms^(-1)`
D. `12ms^(-1)`

Explanation

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

### Step 1: Analyze the first body
- The first body is dropped from height \( h \) with an initial velocity \( u_1 = 0 \, \text{m/s} \).
- It strikes the ground with a final velocity \( v_1 = 3 \, \text{m/s} \).

Using the equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity,
- \( u \) = initial velocity,
- \( a \) = acceleration (which is \( g \), the acceleration due to gravity),
- \( s \) = distance traveled (height \( h \)).

Substituting the known values:
\[ 3^2 = 0^2 + 2gh \]
This simplifies to:
\[ 9 = 2gh \quad \text{(Equation 1)} \]

### Step 2: Analyze the second body
- The second body is also dropped from the same height \( h \) but with an initial velocity \( u_2 = 4 \, \text{m/s} \).
- We need to find the final velocity \( v_2 \) with which it strikes the ground.

Using the same equation of motion:
\[ v_2^2 = u_2^2 + 2gh \]
Substituting the known values:
\[ v_2^2 = 4^2 + 2gh \]
This simplifies to:
\[ v_2^2 = 16 + 2gh \quad \text{(Equation 2)} \]

### Step 3: Substitute \( 2gh \) from Equation 1 into Equation 2
From Equation 1, we know that \( 2gh = 9 \). We can substitute this into Equation 2:
\[ v_2^2 = 16 + 9 \]
This simplifies to:
\[ v_2^2 = 25 \]

### Step 4: Solve for \( v_2 \)
Taking the square root of both sides:
\[ v_2 = \sqrt{25} = 5 \, \text{m/s} \]

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

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