A person sitting in an open car moving at constant velocity throws a ball vertically up into air. The ball falls

Options

A. Outside the car
B. In the car ahead of the person
C. In the car to the side of the person
D. Exactly in the hand which threw it up

Explanation

To solve the problem of a person sitting in an open car moving at constant velocity who throws a ball vertically up into the air, we can analyze the motion of the ball and the car separately.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The car is moving with a constant horizontal velocity \( U \).
- The person throws the ball vertically upward with an initial vertical velocity \( V \).

2. **Analyzing the Vertical Motion of the Ball**:
- When the ball is thrown upwards, it will rise to a maximum height and then fall back down due to gravity.
- The vertical motion of the ball can be described by the equations of motion under constant acceleration (gravity in this case).
- The time taken to reach the maximum height can be calculated using the formula:
\[ t_{up} = \frac{V}{g} \]
- Where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

3. **Calculating the Total Time of Flight**:
- The total time of flight for the ball (up and down) is:
\[ T = 2 \times t_{up} = 2 \times \frac{V}{g} \]

4. **Analyzing the Horizontal Motion**:
- While the ball is in the air, the car continues to move horizontally with velocity \( U \).
- The horizontal distance covered by the car during the time \( T \) is:
\[ D = U \times T = U \times 2 \times \frac{V}{g} \]

5. **Final Position of the Ball**:
- When the ball comes back down, it will not land back in the same spot from where it was thrown because the car has moved forward during the time the ball was in the air.
- The ball will fall a distance \( D \) ahead of the point from which it was thrown.

### Conclusion:
The ball will fall behind the person in the car, at a distance \( D = U \times 2 \times \frac{V}{g} \) from the point where it was thrown.