From the top of a tower of height 100 m, a ball is projected with a velocity of `10 m//"sec"`. It takes 5 seconds to reach the ground. If `g = 10 m//"sec"^(2)` then the angle of projection is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To solve the problem, we need to determine the angle of projection (θ) of a ball that is projected from the top of a tower with a height of 100 meters and an initial velocity of 10 m/s. The ball takes 5 seconds to reach the ground, and we are given that the acceleration due to gravity (g) is 10 m/s².

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Height of the tower (h) = 100 m
- Initial velocity (u) = 10 m/s
- Time taken to reach the ground (t) = 5 s
- Acceleration due to gravity (g) = 10 m/s²

2. **Break Down the Initial Velocity:**
The initial velocity can be resolved into two components:
- Horizontal component (u_x) = u * cos(θ)
- Vertical component (u_y) = u * sin(θ)

So, we have:
- u_x = 10 * cos(θ)
- u_y = 10 * sin(θ)

3. **Set Up the Vertical Motion Equation:**
The vertical displacement (s_y) can be described by the equation of motion:
\[ s_y = u_y \cdot t + \frac{1}{2}(-g) \cdot t^2 \]
Here, the vertical displacement is -100 m (since it falls down), and we substitute the values:
\[ -100 = (10 \sin(θ)) \cdot 5 + \frac{1}{2}(-10) \cdot (5^2) \]

4. **Simplify the Equation:**
Substitute the values:
\[ -100 = 50 \sin(θ) - 125 \]
Rearranging gives:
\[ 50 \sin(θ) = -100 + 125 \]
\[ 50 \sin(θ) = 25 \]
\[ \sin(θ) = \frac{25}{50} = \frac{1}{2} \]

5. **Determine the Angle of Projection:**
The angle θ for which sin(θ) = 1/2 is:
\[ θ = 30^\circ \]

### Conclusion:
The angle of projection is **30 degrees**.