A particle is projected towards the north with speed `20 m//s` at an angle `45^(@)` with horizontal. Ball gets horizontal acceleration of `7.5 m//s^(2)` towards east due to wind. Range of ball (in meter) will be

Answer

To solve the problem of finding the range of a particle projected at an angle with horizontal acceleration, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial speed \( u = 20 \, \text{m/s} \)
- Angle of projection \( \theta = 45^\circ \)
- Horizontal acceleration \( a_x = 7.5 \, \text{m/s}^2 \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

2. **Resolve the Initial Velocity into Components:**
- The horizontal component of the velocity \( u_x \):
\[ u_x = u \cos \theta = 20 \cos 45^\circ = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s} \]
- The vertical component of the velocity \( u_y \):
\[ u_y = u \sin \theta = 20 \sin 45^\circ = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s} \]

3. **Calculate the Time of Flight:**
- The time of flight \( T \) can be calculated using the formula:
\[ T = \frac{2u_y}{g} = \frac{2 \times 10\sqrt{2}}{10} = 2\sqrt{2} \, \text{s} \]

4. **Calculate the Horizontal Range:**
- The horizontal range \( R \) can be calculated using the equation:
\[ R = u_x T + \frac{1}{2} a_x T^2 \]
- Substitute the values:
\[ R = (10\sqrt{2}) \times (2\sqrt{2}) + \frac{1}{2} \times 7.5 \times (2\sqrt{2})^2 \]
- Calculate \( (2\sqrt{2})^2 = 8 \):
\[ R = (10\sqrt{2}) \times (2\sqrt{2}) + \frac{1}{2} \times 7.5 \times 8 \]
- Simplifying further:
\[ R = 20 \times 2 + \frac{1}{2} \times 7.5 \times 8 \]
\[ R = 40 + 30 = 70 \, \text{m} \]

### Final Answer:
The range of the ball is \( R = 70 \, \text{meters} \).