A ball moving is thrown at an angle of `60^(@)` with the horizontal , with initial speed of `30 m//s` . Calculate
(a) time of flight of the ball
(b) hrizontal range .
(c) maximum height attained .

Answer

To solve the problem step by step, we will calculate the time of flight, horizontal range, and maximum height attained by the ball thrown at an angle of 60 degrees with an initial speed of 30 m/s.

### Given Data:
- Initial speed (u) = 30 m/s
- Angle of projection (θ) = 60 degrees
- Acceleration due to gravity (g) = 10 m/s²

### (a) Time of Flight (T)
The formula for the time of flight of a projectile is given by:

\[ T = \frac{2u \sin \theta}{g} \]

**Step 1: Calculate \(\sin 60^\circ\)**
\(\sin 60^\circ = \frac{\sqrt{3}}{2}\)

**Step 2: Substitute values into the formula**
\[ T = \frac{2 \times 30 \times \sin 60^\circ}{10} = \frac{2 \times 30 \times \frac{\sqrt{3}}{2}}{10} \]

**Step 3: Simplify the equation**
\[ T = \frac{30 \sqrt{3}}{10} = 3 \sqrt{3} \]

**Step 4: Calculate the numerical value**
\[ T \approx 3 \times 1.732 \approx 5.196 \text{ seconds} \]

### (b) Horizontal Range (R)
The formula for the horizontal range of a projectile is given by:

\[ R = \frac{u^2 \sin 2\theta}{g} \]

**Step 1: Calculate \(\sin 2\theta\)**
\(\sin 2\theta = \sin 120^\circ = \sin(60^\circ + 60^\circ) = 2 \sin 60^\circ \cos 60^\circ\)
Where \(\cos 60^\circ = \frac{1}{2}\).

Thus,
\(\sin 2\theta = 2 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} = \frac{\sqrt{3}}{2}\)

**Step 2: Substitute values into the formula**
\[ R = \frac{30^2 \times \sin 120^\circ}{10} = \frac{900 \times \frac{\sqrt{3}}{2}}{10} \]

**Step 3: Simplify the equation**
\[ R = \frac{900 \sqrt{3}}{20} = 45 \sqrt{3} \]

**Step 4: Calculate the numerical value**
\[ R \approx 45 \times 1.732 \approx 77.94 \text{ meters} \]

### (c) Maximum Height (H)
The formula for the maximum height attained by a projectile is given by:

\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

**Step 1: Calculate \(\sin^2 60^\circ\)**
\(\sin^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}\)

**Step 2: Substitute values into the formula**
\[ H = \frac{30^2 \times \frac{3}{4}}{2 \times 10} = \frac{900 \times \frac{3}{4}}{20} \]

**Step 3: Simplify the equation**
\[ H = \frac{2700}{80} = 33.75 \text{ meters} \]

### Final Answers:
- (a) Time of Flight = 5.196 seconds
- (b) Horizontal Range = 77.94 meters
- (c) Maximum Height = 33.75 meters