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))`

Explanation

To solve the problem, we need to determine the angle \(\theta\) at which the boy must throw the ball so that it returns to his hand after being affected by a horizontal acceleration due to the wind.

### Step-by-step Solution:

1. **Identify the Components of Initial Velocity:**
The initial velocity \(v_0\) can be broken down into its vertical and horizontal components:
- Vertical component: \(v_{y} = v_0 \sin \theta = 20 \sin \theta\)
- Horizontal component: \(v_{x} = v_0 \cos \theta = 20 \cos \theta\)

2. **Calculate the Time of Flight:**
The time of flight \(t\) for a projectile thrown vertically is given by:
\[ t = \frac{2 v_{y}}{g} = \frac{2 (20 \sin \theta)}{10} = 4 \sin \theta \]

3. **Horizontal Motion Analysis:**
The ball experiences a horizontal acceleration \(a_x = -4 \, \text{m/s}^2\) (to the left). The horizontal displacement \(s_x\) must be zero when the ball returns to the boy's hand. The equation for horizontal motion is:
\[ s_x = v_{x} t + \frac{1}{2} a_x t^2 \]
Setting \(s_x = 0\):
\[ 0 = (20 \cos \theta)(t) + \frac{1}{2}(-4)(t^2) \]

4. **Substituting Time of Flight:**
Substitute \(t = 4 \sin \theta\) into the horizontal motion equation:
\[ 0 = (20 \cos \theta)(4 \sin \theta) - 2(4 \sin \theta)^2 \]
Simplifying:
\[ 0 = 80 \sin \theta \cos \theta - 32 \sin^2 \theta \]

5. **Factor Out \(\sin \theta\):**
Factoring out \(\sin \theta\):
\[ \sin \theta (80 \cos \theta - 32 \sin \theta) = 0 \]
This gives us two cases:
- \(\sin \theta = 0\) (which is not valid as it means the ball is not thrown)
- \(80 \cos \theta - 32 \sin \theta = 0\)

6. **Rearranging the Equation:**
Rearranging the second equation:
\[ 80 \cos \theta = 32 \sin \theta \]
Dividing both sides by \(\cos \theta\):
\[ 80 = 32 \tan \theta \]
Thus:
\[ \tan \theta = \frac{80}{32} = 2.5 \]

7. **Finding the Angle \(\theta\):**
Finally, we find \(\theta\) by taking the arctangent:
\[ \theta = \tan^{-1}(2.5) \]

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