A cricketer can throw a ball to maximum horizontal distance of 160 m. Calculate the maximum vertical height to which he can throw the ball. Given `g=10ms^(-2)`.

Answer

To solve the problem of calculating the maximum vertical height to which a cricketer can throw a ball given the maximum horizontal distance (range) of 160 m, we can follow these steps:

### Step 1: Understand the relationship between range and initial velocity
The formula for the range \( R \) of a projectile is given by:
\[ R = \frac{U^2 \sin 2\theta}{g} \]
where:
- \( R \) is the range,
- \( U \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Step 2: Determine the maximum range
For maximum range, the angle \( \theta \) that gives the maximum range is \( 45^\circ \). Thus, \( \sin 2\theta \) becomes \( \sin 90^\circ = 1 \). Therefore, the formula simplifies to:
\[ R_{\text{max}} = \frac{U^2}{g} \]

### Step 3: Substitute the known values
Given that the maximum horizontal distance \( R_{\text{max}} = 160 \, \text{m} \) and \( g = 10 \, \text{m/s}^2 \), we can write:
\[ 160 = \frac{U^2}{10} \]

### Step 4: Solve for \( U^2 \)
Rearranging the equation to find \( U^2 \):
\[ U^2 = 160 \times 10 = 1600 \]

### Step 5: Calculate the maximum vertical height
The formula for the maximum vertical height \( H \) of a projectile is given by:
\[ H = \frac{U^2 \sin^2 \theta}{2g} \]
For maximum height, we again use \( \theta = 45^\circ \), which gives \( \sin^2 45^\circ = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2} \). Thus, the formula becomes:
\[ H_{\text{max}} = \frac{U^2 \cdot \frac{1}{2}}{2g} = \frac{U^2}{4g} \]

### Step 6: Substitute \( U^2 \) into the height formula
Now substituting \( U^2 = 1600 \) and \( g = 10 \):
\[ H_{\text{max}} = \frac{1600}{4 \times 10} = \frac{1600}{40} = 40 \, \text{m} \]

### Final Answer
The maximum vertical height to which the cricketer can throw the ball is:
\[ \boxed{40 \, \text{m}} \]