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. 50 m
B. 100 m
C. 150 m
D. 200 m

Explanation

To solve the problem of how high a cricketer can throw a ball given that he can throw it to a maximum horizontal distance of 100 meters, we will follow these steps:

### Step 1: Understand the relationship between range and height
The maximum horizontal distance (range) \( R \) for projectile motion is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where:
- \( u \) is the initial speed of the projectile,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 2: Determine the angle for maximum range
The range is maximized when \( \sin 2\theta \) is equal to 1, which occurs when:
\[ 2\theta = 90^\circ \quad \Rightarrow \quad \theta = 45^\circ \]

### Step 3: Calculate the initial speed \( u \)
Using the maximum range formula at \( \theta = 45^\circ \):
\[ R = \frac{u^2}{g} \]
Given that \( R = 100 \, \text{m} \), we can rearrange the equation to find \( u^2 \):
\[ 100 = \frac{u^2}{g} \quad \Rightarrow \quad u^2 = 100g \]

### Step 4: Calculate the height at \( \theta = 45^\circ \)
The maximum height \( H \) reached by the projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Substituting \( \theta = 45^\circ \):
\[ H = \frac{u^2 \sin^2 45^\circ}{2g} \]
Since \( \sin 45^\circ = \frac{1}{\sqrt{2}} \):
\[ H = \frac{u^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

### Step 5: Substitute \( u^2 \) into the height formula
Now, substituting \( u^2 = 100g \) into the height formula:
\[ H = \frac{100g}{4g} = \frac{100}{4} = 25 \, \text{m} \]

### Conclusion
The maximum height above the ground that the cricketer can throw the ball is:
\[ \boxed{25 \, \text{m}} \]

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