A ball is projected horizontally. After 3s from projection its velocity becomes 1.25 times of the velocity of projection. Its velocity of projection is

Options

A. 10 m/s
B. 20 m/s
C. 30 m/s
D. 40 m/s

Answer

To solve the problem, we need to find the velocity of projection of a ball that is projected horizontally. After 3 seconds, its velocity becomes 1.25 times the initial velocity of projection. Let's denote the initial velocity of projection as \( v \).

### Step-by-Step Solution:

1. **Understand the motion**: The ball is projected horizontally, which means its initial vertical velocity (\( u_y \)) is 0 m/s, and it has a horizontal velocity (\( v_x \)) equal to \( v \).

2. **Determine the vertical velocity after 3 seconds**:
The vertical velocity (\( v_y \)) after time \( t \) can be calculated using the equation:
\[ v_y = u_y + a_y \cdot t \]
Here, \( u_y = 0 \), \( a_y = -g \) (where \( g \approx 9.81 \, \text{m/s}^2 \)), and \( t = 3 \, \text{s} \):
\[ v_y = 0 - 9.81 \cdot 3 = -29.43 \, \text{m/s} \]
(We can take the magnitude, so \( v_y = 29.43 \, \text{m/s} \)).

3. **Find the resultant velocity after 3 seconds**:
The resultant velocity (\( V \)) after 3 seconds is given to be 1.25 times the initial velocity of projection:
\[ V = 1.25v \]

4. **Use the Pythagorean theorem to relate the velocities**:
The resultant velocity can be expressed using the horizontal and vertical components:
\[ V = \sqrt{v^2 + v_y^2} \]
Substituting the values we have:
\[ 1.25v = \sqrt{v^2 + (29.43)^2} \]

5. **Square both sides to eliminate the square root**:
\[ (1.25v)^2 = v^2 + (29.43)^2 \]
Expanding this gives:
\[ 1.5625v^2 = v^2 + 867.4249 \]

6. **Rearranging the equation**:
\[ 1.5625v^2 - v^2 = 867.4249 \]
\[ 0.5625v^2 = 867.4249 \]

7. **Solving for \( v^2 \)**:
\[ v^2 = \frac{867.4249}{0.5625} \approx 1540.76 \]

8. **Taking the square root to find \( v \)**:
\[ v = \sqrt{1540.76} \approx 39.24 \, \text{m/s} \]

9. **Final answer**:
Rounding to two decimal places, the velocity of projection is approximately \( 40 \, \text{m/s} \).

### Conclusion:
The velocity of projection is approximately \( 40 \, \text{m/s} \).