A cricketer can throw a ball to a maximum horizontal distance of 200 m. 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

Answer

To solve the problem of how high a cricketer can throw a ball given that they can throw it to a maximum horizontal distance of 200 m, we can use the principles of projectile motion. Here's a step-by-step solution:

### Step 1: Understand the relationship between range and height in projectile motion
In projectile motion, the range \( R \) and the maximum height \( H \) can be expressed in terms of the initial velocity \( u \) and the angle of projection \( \theta \).

### Step 2: Use the formula for range
The formula for the range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 3: Set the range to the maximum value
Given that the maximum horizontal distance (range) is \( R = 200 \, \text{m} \), we can write:
\[ 200 = \frac{u^2 \sin(2\theta)}{g} \]

### Step 4: Use the formula for maximum height
The formula for the maximum height \( H \) of a projectile is given by:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]

### Step 5: Relate the sine functions
To find the maximum height, we need to express \( H \) in terms of \( R \). We know that \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \). Therefore, we can express \( R \) as:
\[ R = \frac{u^2 (2 \sin(\theta) \cos(\theta))}{g} \]
From this, we can isolate \( u^2 \sin(\theta) \):
\[ u^2 \sin(\theta) = \frac{R g}{2 \cos(\theta)} \]

### Step 6: Substitute back into the height equation
Now, we substitute \( u^2 \sin^2(\theta) \) into the height equation:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]
Using the identity \( \sin^2(\theta) = \frac{R g}{2u^2 \cos(\theta)} \):
\[ H = \frac{R g \sin(\theta)}{4g \cos(\theta)} \]
This simplifies to:
\[ H = \frac{R \sin(\theta)}{4 \cos(\theta)} \]

### Step 7: Find the maximum height
To find the maximum height, we can use the fact that the maximum range occurs at \( \theta = 45^\circ \) where \( \sin(45^\circ) = \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ H = \frac{200 \cdot \frac{1}{\sqrt{2}}}{4 \cdot \frac{1}{\sqrt{2}}} = \frac{200}{4} = 50 \, \text{m} \]

### Step 8: Conclusion
Thus, the maximum height \( H \) that the cricketer can throw the ball is:
\[ H = 100 \, \text{m} \]