A cricketer hits a ball with a velocity `25m//s` at `60^(@)` above the horizontal. How far above the ground it passes over a fielder 50 m from the bat (assume the ball is struck very close to the ground)

Options

A. `6.60 m`
B. `9.0 m`
C. `11.6 m`
D. `12.7 m`

Answer

To solve the problem of how far above the ground the ball passes over a fielder 50 m from the bat, we will follow these steps:

### Step 1: Identify the given parameters
- Initial velocity of the ball, \( u = 25 \, \text{m/s} \)
- Angle of projection, \( \theta = 60^\circ \)
- Horizontal distance to the fielder, \( x = 50 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the horizontal and vertical components of the initial velocity
- The horizontal component of the velocity:
\[ u_x = u \cos \theta = 25 \cos 60^\circ = 25 \times \frac{1}{2} = 12.5 \, \text{m/s} \]
- The vertical component of the velocity:
\[ u_y = u \sin \theta = 25 \sin 60^\circ = 25 \times \frac{\sqrt{3}}{2} \approx 21.65 \, \text{m/s} \]

### Step 3: Calculate the time of flight to reach the fielder
Using the horizontal motion equation:
\[ x = u_x t \implies t = \frac{x}{u_x} = \frac{50}{12.5} = 4 \, \text{s} \]

### Step 4: Calculate the vertical position of the ball when it reaches the fielder
Using the vertical motion equation:
\[ y = u_y t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ y = (21.65 \times 4) - \frac{1}{2} \times 10 \times (4^2) \]
Calculating each term:
\[ y = 86.6 - \frac{1}{2} \times 10 \times 16 \]
\[ y = 86.6 - 80 \]
\[ y = 6.6 \, \text{m} \]

### Conclusion
The ball passes over the fielder at a height of **6.6 meters** above the ground.

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