A boy throws a ball upward with velocity `v_(0)=20m//s`. The wind imparts a horizontal acceleration of `4m//s^(2)` to the left. The angle `theta` at which the ball must be thrown so that the ball returns to the boy's hand is : `(g=10ms^(-2))`

Options

A. `tan^(-1)(1.2)`
B. `tan^(-1)(0.2)`
C. `tan^(-1)(2)`
D. `tan^(-1)(0.4)`

Explanation

To solve the problem step by step, we can break it down as follows:

### Step 1: Understand the Problem
We have a boy throwing a ball upward with an initial velocity \( v_0 = 20 \, \text{m/s} \). The ball experiences a horizontal acceleration due to the wind of \( a_x = 4 \, \text{m/s}^2 \) to the left. We need to find the angle \( \theta \) at which the ball should be thrown so that it returns to the boy's hand.

### Step 2: Set Up the Equations
1. **Vertical Motion**:
- The vertical component of the initial velocity is given by:
\[ v_{y0} = v_0 \sin \theta \]
- The time of flight \( t \) can be calculated using the formula for vertical motion under gravity:
\[ t = \frac{2 v_{y0}}{g} = \frac{2 v_0 \sin \theta}{g} \]
- Here, \( g = 10 \, \text{m/s}^2 \).

2. **Horizontal Motion**:
- The horizontal component of the initial velocity is given by:
\[ v_{x0} = v_0 \cos \theta \]
- The horizontal displacement must be zero when the ball returns to the boy's hand. The horizontal motion is described by:
\[ x = v_{x0} t + \frac{1}{2} a_x t^2 \]
- Substituting \( a_x = -4 \, \text{m/s}^2 \) (since it's to the left):
\[ 0 = v_0 \cos \theta \cdot t - 2 t^2 \]

### Step 3: Solve for Time \( t \)
From the horizontal motion equation:
\[ v_0 \cos \theta \cdot t = 2 t^2 \]
Assuming \( t \neq 0 \), we can divide both sides by \( t \):
\[ v_0 \cos \theta = 2 t \]
Thus,
\[ t = \frac{v_0 \cos \theta}{2} \]

### Step 4: Equate the Two Expressions for Time \( t \)
From the vertical motion, we have:
\[ t = \frac{2 v_0 \sin \theta}{g} \]
Setting the two expressions for \( t \) equal to each other:
\[ \frac{v_0 \cos \theta}{2} = \frac{2 v_0 \sin \theta}{g} \]
Cancelling \( v_0 \) (since \( v_0 \neq 0 \)):
\[ \frac{\cos \theta}{2} = \frac{2 \sin \theta}{g} \]
Substituting \( g = 10 \):
\[ \frac{\cos \theta}{2} = \frac{2 \sin \theta}{10} \]
This simplifies to:
\[ \cos \theta = \frac{4 \sin \theta}{10} = \frac{2 \sin \theta}{5} \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ \frac{\cos \theta}{\sin \theta} = \frac{2}{5} \]
Thus,
\[ \tan \theta = \frac{5}{2} \]

### Step 6: Calculate \( \theta \)
Now, we find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{5}{2}\right) \]

### Final Answer
The angle \( \theta \) at which the ball must be thrown so that it returns to the boy's hand is:
\[ \theta = \tan^{-1}(2.5) \]