A ball is thrown from the top of a tower with an initial velocity of `10 m//s` at an angle of `30^(@)`above the horizontal. It hits the ground at a distance of 17.3 m from the base of the tower. The height of the tower `(g=10m//s^(2))` will be

Options

A. 5 m
B. 20 m
C. 15 m
D. 10 m

Explanation

To solve the problem step by step, we will break down the motion of the ball into its horizontal and vertical components and use the equations of motion.

### Step 1: Identify the given values
- Initial velocity, \( u = 10 \, \text{m/s} \)
- Angle of projection, \( \theta = 30^\circ \)
- Horizontal distance from the base of the tower, \( R = 17.3 \, \text{m} \)
- 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, \( u_x = u \cos \theta \)
\[ u_x = 10 \cos(30^\circ) = 10 \times \frac{\sqrt{3}}{2} = 10 \times 0.866 \approx 8.66 \, \text{m/s} \]

- The vertical component of the velocity, \( u_y = u \sin \theta \)
\[ u_y = 10 \sin(30^\circ) = 10 \times \frac{1}{2} = 5 \, \text{m/s} \]

### Step 3: Calculate the time of flight using horizontal motion
The horizontal distance traveled is given by:
\[ R = u_x \cdot t \]
Rearranging for time \( t \):
\[ t = \frac{R}{u_x} = \frac{17.3}{8.66} \approx 2 \, \text{s} \]

### Step 4: Calculate the vertical displacement using vertical motion
Using the vertical motion equation:
\[ h = u_y \cdot t - \frac{1}{2} g t^2 \]
Substituting the known values:
\[ h = 5 \cdot 2 - \frac{1}{2} \cdot 10 \cdot (2^2) \]
Calculating each term:
\[ h = 10 - \frac{1}{2} \cdot 10 \cdot 4 = 10 - 20 = -10 \, \text{m} \]

### Step 5: Interpret the result
The negative sign indicates that the ball has fallen 10 meters below the point of projection, which means the height of the tower is:
\[ \text{Height of the tower} = 10 \, \text{m} \]

### Final Answer
The height of the tower is \( 10 \, \text{m} \).

---