A cricket ball is thrown at a speed of ` 28 ms^(-1)` in a direction `30 ^(@)` above the horizontal. Calculate (a)the maximum height (b) the time taken by ball to return to the same level, and (c )the distance from the thrower to the point where the ball returns to the same level.

Options

A. ` 10.0m` `2.8 s` `29 m`
B. ` 10.0m` `2.8 s` `69 m`
C. ` 10.0m` `3.8 s` `69 m`
D. ` 20.0m` `2.8 s` `69 m`

Explanation

To solve the problem step by step, we will break down each part of the question.

### Given Data:
- Initial speed (u) = 28 m/s
- Angle of projection (θ) = 30°
- Acceleration due to gravity (g) = 10 m/s²

### (a) Calculate the Maximum Height (H)

1. **Determine the vertical component of the initial velocity (u_y)**:
\[ u_y = u \cdot \sin(\theta) = 28 \cdot \sin(30°) = 28 \cdot \frac{1}{2} = 14 \, \text{m/s} \]

2. **Use the formula for maximum height (H)**:
\[ H = \frac{u_y^2}{2g} = \frac{(14)^2}{2 \cdot 10} = \frac{196}{20} = 9.8 \, \text{m} \]

### (b) Calculate the Time Taken by the Ball to Return to the Same Level (T)

1. **Use the formula for time of flight (T)**:
\[ T = \frac{2u_y}{g} = \frac{2 \cdot 14}{10} = \frac{28}{10} = 2.8 \, \text{s} \]

### (c) Calculate the Horizontal Distance (Range, R)

1. **Determine the horizontal component of the initial velocity (u_x)**:
\[ u_x = u \cdot \cos(\theta) = 28 \cdot \cos(30°) = 28 \cdot \frac{\sqrt{3}}{2} = 14\sqrt{3} \, \text{m/s} \]

2. **Use the formula for range (R)**:
\[ R = u_x \cdot T = 14\sqrt{3} \cdot 2.8 \]
First, calculate \(14\sqrt{3}\):
\[ \sqrt{3} \approx 1.732 \implies 14\sqrt{3} \approx 14 \cdot 1.732 \approx 24.248 \, \text{m/s} \]
Now calculate the range:
\[ R \approx 24.248 \cdot 2.8 \approx 67.9 \, \text{m} \]

### Final Answers:
- (a) Maximum Height (H) = 9.8 m
- (b) Time Taken (T) = 2.8 s
- (c) Distance (R) = 67.9 m

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