A boy throws a ball at an angle `theta` with the vertical. If the vertical component of the initial velocity is `20ms^(-1)` and the wind imparts a horizontal acceleration of `8ms^(-2)` to the left, the angle at which the ball must be thrown so that the ball returns to the boy's hand is `theta`. What is the value of `10(tan theta)`

Answer

To solve the problem, we need to analyze the motion of the ball thrown by the boy at an angle \( \theta \) with the vertical. We know the vertical component of the initial velocity and the horizontal acceleration due to wind. Let's break it down step by step.

### Step 1: Identify the components of the initial velocity
The vertical component of the initial velocity is given as:
\[ V_y = 20 \, \text{m/s} \]
Let the initial velocity of the ball be \( V \). The vertical and horizontal components can be expressed as:
\[ V_y = V \cos(\theta) \quad \text{(vertical component)} \]
\[ V_x = V \sin(\theta) \quad \text{(horizontal component)} \]

### Step 2: Determine the time of flight
The time of flight \( t \) can be calculated using the vertical motion equation. The ball returns to the same vertical position after time \( t \), so we can use the equation:
\[ 0 = V_y t - \frac{1}{2} g t^2 \]
Substituting \( V_y = 20 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ 0 = 20t - \frac{1}{2} \cdot 10 \cdot t^2 \]
\[ 0 = 20t - 5t^2 \]
Factoring out \( t \):
\[ t(20 - 5t) = 0 \]
This gives us two solutions: \( t = 0 \) (initial time) or \( t = \frac{20}{5} = 4 \, \text{s} \).

### Step 3: Calculate horizontal displacement
The horizontal displacement \( x \) when the ball returns to the boy can be calculated using the horizontal motion equation:
\[ x = V_x t + \frac{1}{2} a_x t^2 \]
Given that the horizontal acceleration \( a_x = -8 \, \text{m/s}^2 \) (to the left):
\[ x = V_x t - 4t^2 \]
Substituting \( t = 4 \, \text{s} \):
\[ x = V_x \cdot 4 - 4 \cdot 4^2 \]
\[ x = V_x \cdot 4 - 64 \]

### Step 4: Relate horizontal and vertical components
Since the ball returns to the boy's hand, the horizontal displacement must be zero:
\[ 0 = V_x \cdot 4 - 64 \]
Thus, we can solve for \( V_x \):
\[ V_x \cdot 4 = 64 \implies V_x = 16 \, \text{m/s} \]

### Step 5: Relate \( V_x \) and \( V_y \) to find \( \tan(\theta) \)
Now we have:
\[ V_x = V \sin(\theta) = 16 \quad \text{and} \quad V_y = V \cos(\theta) = 20 \]
We can find \( \tan(\theta) \):
\[ \tan(\theta) = \frac{V_y}{V_x} = \frac{20}{16} = \frac{5}{4} \]

### Step 6: Calculate \( 10 \tan(\theta) \)
Finally, we calculate:
\[ 10 \tan(\theta) = 10 \cdot \frac{5}{4} = 12.5 \]

### Final Answer
The value of \( 10 \tan(\theta) \) is:
\[ \boxed{12.5} \]