A man is riding on a flat car travelling with a constant speed of `10m//s`.He wishes to throw a ball through a stationary hoop `15 m` above the height of his hands in such a manner that the ball will move horizontally as it passes through the hoop. He throws the ball with a speed of `12.5 m//s` w.r.t himself.
At what horizontal distance in front of the hoop must he release the ball?

Options

A. `12.5 m`
B. `15.5 m`
C. `17.5 m`
D. `20 m`

Explanation

To solve the problem, we need to determine the horizontal distance in front of the hoop where the man must release the ball so that it passes through the hoop horizontally.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Speed of the car (v_car) = 10 m/s
- Speed of the ball with respect to the man (v_ball) = 12.5 m/s
- Height of the hoop above the man's hands (h) = 15 m
- The ball must pass through the hoop horizontally.

2. **Determine the Vertical Motion:**
- The ball is thrown at an angle θ with respect to the horizontal.
- The vertical component of the ball's velocity (v_vertical) is given by:
\[ v_{vertical} = v_{ball} \cdot \sin(\theta) \]
- The horizontal component of the ball's velocity (v_horizontal) is given by:
\[ v_{horizontal} = v_{ball} \cdot \cos(\theta) + v_{car} \]

3. **Use the Vertical Motion Equation:**
- Since the ball passes through the hoop horizontally, its vertical velocity at that point is zero.
- Using the kinematic equation for vertical motion:
\[ h = v_{vertical} \cdot t - \frac{1}{2} g t^2 \]
- Here, \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).

4. **Calculate Time to Reach the Hoop:**
- The ball must fall a vertical distance of 15 m, so we set up the equation:
\[ 15 = v_{ball} \cdot \sin(\theta) \cdot t - \frac{1}{2} g t^2 \]
- Rearranging gives us a quadratic equation in terms of \( t \):
\[ \frac{1}{2} g t^2 - v_{ball} \cdot \sin(\theta) \cdot t + 15 = 0 \]

5. **Solve for Time (t):**
- Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we can find \( t \).

6. **Calculate Horizontal Distance:**
- The horizontal distance (d) traveled by the ball when it reaches the hoop is given by:
\[ d = v_{horizontal} \cdot t \]
- Substitute \( v_{horizontal} = v_{ball} \cdot \cos(\theta) + v_{car} \) into the equation.

7. **Determine the Angle θ:**
- From the vertical motion, we can find \( \sin(\theta) \) and \( \cos(\theta) \) using trigonometric identities based on the relationship between the vertical and horizontal components.

8. **Final Calculation:**
- Substitute the values into the equations to find the horizontal distance in front of the hoop.