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. The ball is thrown at an angle \( \theta \) with the vertical, and we have the following information:

- The vertical component of the initial velocity \( u_y = 20 \, \text{m/s} \).
- The horizontal acceleration due to wind \( a_x = -8 \, \text{m/s}^2 \) (to the left).

### Step 1: Determine the time of flight

Since the ball returns to the boy's hand, the net vertical displacement is zero. We can use the equation of motion for vertical displacement:

\[ y = u_y t + \frac{1}{2} a_y t^2 \]

Here, \( a_y = -g = -10 \, \text{m/s}^2 \) (acceleration due to gravity), and \( y = 0 \) (net vertical displacement). Thus, we have:

\[ 0 = 20t - \frac{1}{2} \cdot 10 \cdot t^2 \]

This simplifies to:

\[ 0 = 20t - 5t^2 \]

Factoring out \( t \):

\[ t(20 - 5t) = 0 \]

This gives us two solutions: \( t = 0 \) (initial time) and \( t = 4 \, \text{s} \) (time of flight).

### Step 2: Determine the horizontal displacement

The horizontal displacement must also be zero, so we use the equation for horizontal motion:

\[ x = u_x t + \frac{1}{2} a_x t^2 \]

Where \( u_x = u \sin \theta \) and \( a_x = -8 \, \text{m/s}^2 \). Setting \( x = 0 \):

\[ 0 = u \sin \theta \cdot 4 - \frac{1}{2} \cdot 8 \cdot (4^2) \]

This simplifies to:

\[ 0 = 4u \sin \theta - 64 \]

Rearranging gives:

\[ 4u \sin \theta = 64 \]

Dividing by 4:

\[ u \sin \theta = 16 \quad \text{(Equation 1)} \]

### Step 3: Relate vertical and horizontal components

We also know from the problem that:

\[ u \cos \theta = 20 \quad \text{(Equation 2)} \]

### Step 4: Find \( \tan \theta \)

Now, we can find \( \tan \theta \) by dividing Equation 1 by Equation 2:

\[ \frac{u \sin \theta}{u \cos \theta} = \frac{16}{20} \]

This simplifies to:

\[ \tan \theta = \frac{16}{20} = \frac{4}{5} \]

### Step 5: Calculate \( 10 \tan \theta \)

Finally, we need to find \( 10 \tan \theta \):

\[ 10 \tan \theta = 10 \cdot \frac{4}{5} = 8 \]

Thus, the value of \( 10(tan \theta) \) is \( \boxed{8} \).

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