A body is projected horizontally from the top of a tower with a velocity of 10m/s. If it hits the ground at an angle of `45^@`, the vertical component of velocity when it hits ground in m/s is 

Options

A. `10 sqrt(2)`
B. `5sqrt(2)`
C. `5`
D. `10`

Answer

To solve the problem, we need to determine the vertical component of the velocity of a body projected horizontally from the top of a tower when it hits the ground at an angle of 45 degrees.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The body is projected horizontally with an initial horizontal velocity \( u_x = 10 \, \text{m/s} \).
- The initial vertical velocity \( u_y = 0 \, \text{m/s} \) since it is projected horizontally.

2. **Understanding the Motion:**
- The body will fall under the influence of gravity, which means there will be a vertical acceleration \( a_y = g \) (where \( g \approx 9.81 \, \text{m/s}^2 \)).
- The horizontal velocity remains constant because there is no horizontal acceleration. Thus, \( v_x = u_x = 10 \, \text{m/s} \).

3. **Using the Angle of Impact:**
- When the body hits the ground at an angle of \( 45^\circ \), the relationship between the vertical and horizontal components of the velocity can be expressed as:
\[ \tan(45^\circ) = \frac{v_y}{v_x} \]
- Since \( \tan(45^\circ) = 1 \), we have:
\[ v_y = v_x \]

4. **Substituting the Known Values:**
- From the previous steps, we know \( v_x = 10 \, \text{m/s} \). Therefore:
\[ v_y = 10 \, \text{m/s} \]

5. **Conclusion:**
- The vertical component of the velocity when the body hits the ground is \( v_y = 10 \, \text{m/s} \).

### Final Answer:
The vertical component of the velocity when it hits the ground is \( 10 \, \text{m/s} \).

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