A cricketer can throw a ball to a maximum horizontal distance of 100m. With the same speed how much high above the ground can the cricketer throw the same ball?

Options

A. `150 m`
B. `250 m`
C. `350 m`
D. `50 m`

Explanation

To solve the problem, we need to determine how high a cricketer can throw a ball when it is thrown with the same speed used to achieve a maximum horizontal distance of 100 meters.

### Step-by-Step Solution:

1. **Understanding the Maximum Range**:
The maximum horizontal distance (range) for projectile motion is achieved when the projectile is thrown at an angle of 45 degrees. The formula for the range \( R \) is given by:
\[ R = \frac{v_0^2 \sin(2\theta)}{g} \]
where \( v_0 \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

2. **Substituting Values**:
Since the angle \( \theta = 45^\circ \), we have:
\[ \sin(90^\circ) = 1 \]
Therefore, the formula simplifies to:
\[ R = \frac{v_0^2}{g} \]
Given that the range \( R = 100 \, \text{m} \), we can set up the equation:
\[ 100 = \frac{v_0^2}{10} \]

3. **Solving for Initial Velocity \( v_0 \)**:
Rearranging the equation to solve for \( v_0^2 \):
\[ v_0^2 = 1000 \]
Taking the square root gives:
\[ v_0 = 10\sqrt{10} \, \text{m/s} \]

4. **Calculating Maximum Height**:
The maximum height \( h_{\text{max}} \) that can be achieved when throwing the ball vertically is given by the formula:
\[ h_{\text{max}} = \frac{v_0^2}{2g} \]
Substituting \( v_0^2 = 1000 \) and \( g = 10 \):
\[ h_{\text{max}} = \frac{1000}{2 \times 10} = \frac{1000}{20} = 50 \, \text{m} \]

5. **Final Answer**:
The maximum height above the ground that the cricketer can throw the ball is:
\[ h_{\text{max}} = 50 \, \text{m} \]