The maximum distance to which a man can throw a ball by projecting it horizontally from a height h is h. The maximum distance to which he can throw it vertically up is

Options

A. h
B. 2h
C. h/2
D. h/4

Answer

To solve the problem, we need to find the maximum height to which a man can throw a ball vertically upward from a height \( h \). We know that when the ball is thrown horizontally from height \( h \), it travels a horizontal distance \( h \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- When a ball is thrown horizontally from a height \( h \), it takes time \( t \) to hit the ground. This time can be calculated using the equation of motion for free fall.

2. **Calculate the Time of Flight**:
- The distance fallen under gravity is given by:
\[ h = \frac{1}{2} g t^2 \]
- Rearranging this gives:
\[ t^2 = \frac{2h}{g} \]
- Taking the square root:
\[ t = \sqrt{\frac{2h}{g}} \]

3. **Horizontal Velocity**:
- The horizontal distance traveled is equal to the horizontal velocity multiplied by the time of flight. Since the horizontal distance is also \( h \), we can set up the equation:
\[ h = u \cdot t \]
- Substituting for \( t \):
\[ h = u \cdot \sqrt{\frac{2h}{g}} \]
- Solving for \( u \):
\[ u = \frac{h}{\sqrt{\frac{2h}{g}}} = \sqrt{\frac{gh}{2}} \]

4. **Throwing the Ball Vertically Upward**:
- Now, we will use the velocity \( u \) to find the maximum height when the ball is thrown vertically upward. At the maximum height, the final velocity \( v = 0 \).
- Using the third equation of motion:
\[ v^2 = u^2 - 2gh' \]
- Setting \( v = 0 \):
\[ 0 = u^2 - 2gh' \]
- Rearranging gives:
\[ h' = \frac{u^2}{2g} \]

5. **Substituting for \( u^2 \)**:
- We already found \( u = \sqrt{\frac{gh}{2}} \), so:
\[ u^2 = \frac{gh}{2} \]
- Substituting this into the equation for \( h' \):
\[ h' = \frac{\frac{gh}{2}}{2g} = \frac{h}{4} \]

6. **Final Answer**:
- Therefore, the maximum height to which the man can throw the ball vertically upward is:
\[ \boxed{\frac{h}{4}} \]