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)`

Answer

To solve the problem, we need to find the angle \(\theta\) at which the boy must throw the ball so that it returns to his hand, considering the horizontal acceleration due to wind. Let's break this down step by step.

### Step 1: Identify the components of the initial velocity
The initial velocity \(v_0\) can be resolved into horizontal and vertical components:
- Vertical component: \(v_y = v_0 \sin \theta\)
- Horizontal component: \(v_x = v_0 \cos \theta\)

### Step 2: Determine the time of flight
The time of flight \(T\) for a projectile thrown vertically with an initial vertical velocity \(v_y\) and under the influence of gravity \(g\) is given by:
\[ T = \frac{2v_y}{g} = \frac{2(v_0 \sin \theta)}{g} \]

### Step 3: Analyze horizontal motion
The horizontal motion is affected by the wind, which imparts a horizontal acceleration \(a_x = -4 \, \text{m/s}^2\) (to the left). The horizontal displacement \(x\) during the time of flight must be zero for the ball to return to the boy's hand. The equation for horizontal displacement is:
\[ x = v_x T + \frac{1}{2} a_x T^2 \]
Substituting \(x = 0\):
\[ 0 = v_0 \cos \theta \cdot T + \frac{1}{2} (-4) T^2 \]

### Step 4: Substitute the time of flight into the horizontal motion equation
Substituting \(T\) from step 2 into the horizontal motion equation:
\[ 0 = v_0 \cos \theta \cdot \left(\frac{2(v_0 \sin \theta)}{g}\right) - 2 T^2 \]
This simplifies to:
\[ 0 = \frac{2v_0^2 \sin \theta \cos \theta}{g} - 2 \left(\frac{2(v_0 \sin \theta)}{g}\right)^2 \]

### Step 5: Simplify the equation
Rearranging gives:
\[ \frac{2v_0^2 \sin \theta \cos \theta}{g} = \frac{8(v_0^2 \sin^2 \theta)}{g^2} \]
Cancelling \(2v_0^2\) from both sides (assuming \(v_0 \neq 0\)):
\[ \sin \theta \cos \theta = \frac{4 \sin^2 \theta}{g} \]

### Step 6: Use the identity \(\sin(2\theta) = 2 \sin \theta \cos \theta\)
This gives:
\[ \frac{1}{2} \sin(2\theta) = \frac{4 \sin^2 \theta}{g} \]
Substituting \(g = 10 \, \text{m/s}^2\):
\[ \frac{1}{2} \sin(2\theta) = \frac{4 \sin^2 \theta}{10} \]
This simplifies to:
\[ \sin(2\theta) = \frac{8 \sin^2 \theta}{10} \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ 10 \sin(2\theta) = 8 \sin^2 \theta \]
Using the identity \(\sin(2\theta) = 2 \sin \theta \cos \theta\):
\[ 10 \cdot 2 \sin \theta \cos \theta = 8 \sin^2 \theta \]
This simplifies to:
\[ 20 \sin \theta \cos \theta = 8 \sin^2 \theta \]

### Step 8: Dividing through by \(\sin \theta\) (assuming \(\sin \theta \neq 0\))
\[ 20 \cos \theta = 8 \sin \theta \]
This can be rearranged to:
\[ \tan \theta = \frac{20}{8} = 2.5 \]

### Step 9: Finding the angle \(\theta\)
Finally, we can find \(\theta\):
\[ \theta = \tan^{-1}(2.5) \]

### Final Answer
Thus, the angle \(\theta\) at which the ball must be thrown so that it returns to the boy's hand is:
\[ \theta \approx 68.2^\circ \]