A ball is thrown at a speed of 40 m/s at an angle of `60^0` with the horizontal. Find a. the maximum height reached and b. the range of te ball. Take `g=10 m/s^2`.

Explanation

To solve the problem of a ball thrown at a speed of 40 m/s at an angle of 60 degrees with the horizontal, we will find the maximum height reached by the ball and the range of the ball.

### Step-by-Step Solution:

**Given:**
- Initial speed (u) = 40 m/s
- Angle of projection (θ) = 60 degrees
- Acceleration due to gravity (g) = 10 m/s²

**a. To find the maximum height (H_max):**

1. **Use the formula for maximum height:**
\[ H_{max} = \frac{u^2 \sin^2 \theta}{2g} \]

2. **Calculate \( \sin 60^\circ \):**
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]

3. **Substitute the values into the formula:**
\[ H_{max} = \frac{(40)^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2 \times 10} \]

4. **Calculate \( (40)^2 \):**
\[ (40)^2 = 1600 \]

5. **Calculate \( \left(\frac{\sqrt{3}}{2}\right)^2 \):**
\[ \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

6. **Now substitute back into the equation:**
\[ H_{max} = \frac{1600 \times \frac{3}{4}}{20} \]

7. **Simplify the equation:**
\[ H_{max} = \frac{1600 \times 3}{80} = \frac{4800}{80} = 60 \text{ meters} \]

**b. To find the range (R):**

1. **Use the formula for range:**
\[ R = \frac{u^2 \sin 2\theta}{g} \]

2. **Calculate \( \sin 2\theta \):**
\[ \sin 2\theta = \sin 120^\circ = \frac{\sqrt{3}}{2} \]

3. **Substitute the values into the formula:**
\[ R = \frac{(40)^2 \cdot \frac{\sqrt{3}}{2}}{10} \]

4. **Calculate \( (40)^2 \):**
\[ (40)^2 = 1600 \]

5. **Now substitute back into the equation:**
\[ R = \frac{1600 \cdot \frac{\sqrt{3}}{2}}{10} \]

6. **Simplify the equation:**
\[ R = \frac{1600 \cdot \sqrt{3}}{20} = 80\sqrt{3} \text{ meters} \]

### Final Answers:
- Maximum Height (H_max) = 60 meters
- Range (R) = 80√3 meters