A cricketer of height `2.5 m` thrown a ball at an angle of `30^(@)` with the horizontal such that it is received by another crickerter of same height standing at distance of `50m` from the first one.Find the maximum height attained by the ball.

Options

A. `10 m`
B. `9 m`
C. `10.7 m`
D. `9.7 m`

Explanation

To solve the problem of finding the maximum height attained by the ball thrown by the cricketer, we can follow these steps:

### Step 1: Understand the Problem
We have a cricketer throwing a ball at an angle of \(30^\circ\) with the horizontal. The ball is caught by another cricketer of the same height (2.5 m) standing 50 m away. We need to find the maximum height attained by the ball above the ground.

### Step 2: Identify the Variables
- Height of the cricketers: \(h = 2.5 \, \text{m}\)
- Distance between the cricketers: \(R = 50 \, \text{m}\)
- Angle of projection: \(\theta = 30^\circ\)
- Acceleration due to gravity: \(g = 9.81 \, \text{m/s}^2\)

### Step 3: Use the Range Formula
The range \(R\) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Where \(u\) is the initial velocity of the ball. We can rearrange this to find \(u^2\):
\[ u^2 = \frac{R \cdot g}{\sin(2\theta)} \]

### Step 4: Calculate \(\sin(2\theta)\)
For \(\theta = 30^\circ\):
\[ \sin(2\theta) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

### Step 5: Substitute Values into the Range Formula
Substituting the known values into the range formula:
\[ u^2 = \frac{50 \cdot 9.81}{\frac{\sqrt{3}}{2}} = \frac{50 \cdot 9.81 \cdot 2}{\sqrt{3}} = \frac{981}{\sqrt{3}} \approx 567.1 \, \text{m}^2/\text{s}^2 \]

### Step 6: Calculate Maximum Height
The maximum height \(h_{max}\) attained by the projectile is given by:
\[ h_{max} = \frac{u^2 \sin^2(\theta)}{2g} \]
Substituting \(\sin(30^\circ) = \frac{1}{2}\):
\[ h_{max} = \frac{567.1 \cdot \left(\frac{1}{2}\right)^2}{2 \cdot 9.81} = \frac{567.1 \cdot \frac{1}{4}}{19.62} = \frac{141.775}{19.62} \approx 7.22 \, \text{m} \]

### Step 7: Calculate Total Height
The total height above the ground is:
\[ \text{Total Height} = h_{max} + h = 7.22 + 2.5 = 9.72 \, \text{m} \]

### Final Answer
The maximum height attained by the ball above the ground is approximately **9.72 m**.

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