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) minus 42 m will be

Answer

To solve the problem, we need to find the range of the particle projected at an angle with an initial speed and with an additional horizontal acceleration due to wind. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the initial velocity components
The particle is projected with a speed of \(20 \, \text{m/s}\) at an angle of \(45^\circ\) with the horizontal. We can resolve this velocity into its horizontal and vertical components.

- Horizontal component (\(u_x\)):
\[ u_x = u \cdot \cos(45^\circ) = 20 \cdot \frac{1}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s} \]

- Vertical component (\(u_y\)):
\[ u_y = u \cdot \sin(45^\circ) = 20 \cdot \frac{1}{\sqrt{2}} = 10\sqrt{2} \, \text{m/s} \]

### Step 2: Calculate the time of flight
The time of flight can be calculated using the formula:
\[ t = \frac{2u_y}{g} \]
where \(g\) is the acceleration due to gravity (\(g \approx 10 \, \text{m/s}^2\)).

Substituting the values:
\[ t = \frac{2 \cdot 10\sqrt{2}}{10} = 2\sqrt{2} \, \text{s} \]

### Step 3: Calculate the horizontal range
The horizontal range \(R\) can be calculated using the formula:
\[ R = u_x \cdot t + \frac{1}{2} a_x t^2 \]
where \(a_x\) is the horizontal acceleration due to wind (\(7.5 \, \text{m/s}^2\)).

Substituting the values:
\[ R = (10\sqrt{2}) \cdot (2\sqrt{2}) + \frac{1}{2} \cdot 7.5 \cdot (2\sqrt{2})^2 \]

Calculating each term:
1. First term:
\[ (10\sqrt{2}) \cdot (2\sqrt{2}) = 20 \cdot 2 = 40 \, \text{m} \]

2. Second term:
\[ \frac{1}{2} \cdot 7.5 \cdot (2\sqrt{2})^2 = \frac{1}{2} \cdot 7.5 \cdot 8 = 30 \, \text{m} \]

Adding both terms together:
\[ R = 40 + 30 = 70 \, \text{m} \]

### Step 4: Find the final answer
The problem asks for the range of the ball minus 42 m:
\[ R - 42 = 70 - 42 = 28 \, \text{m} \]

Thus, the final answer is:
\[ \text{Range of ball (in meter) minus 42 m will be } 28 \, \text{m}. \]