A ball is thrown from the top of tower with an initial velocity of `10ms^(-1)` at an angle of `30^(@)` with the horizontal. If it hits the ground of a distance of 17.3m from the back of the tower, the height of the tower is `(Take g=10ms^(-2))`

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Break down the initial velocity into components
The initial velocity \( u = 10 \, \text{m/s} \) is thrown at an angle of \( 30^\circ \) with the horizontal. We can find the horizontal and vertical components of the velocity using trigonometric functions:

- Horizontal component \( u_x = u \cos \theta = 10 \cos 30^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \)
- Vertical component \( u_y = u \sin \theta = 10 \sin 30^\circ = 10 \times \frac{1}{2} = 5 \, \text{m/s} \)

### Step 2: Calculate the time of flight
The horizontal distance covered by the ball is given as \( 17.3 \, \text{m} \). We can use the horizontal motion to find the time of flight \( t \):

\[ \text{Distance} = \text{Velocity} \times \text{Time} \implies t = \frac{\text{Distance}}{u_x} = \frac{17.3}{5\sqrt{3}} \]

Calculating \( t \):

\[ t = \frac{17.3}{5 \times 1.732} \approx \frac{17.3}{8.66} \approx 2 \, \text{s} \]

### Step 3: Use the vertical motion to find the height of the tower
Now we can use the vertical motion to find the height \( h \) of the tower. The vertical displacement can be calculated using the second equation of motion:

\[ h = u_y t - \frac{1}{2} g t^2 \]

Substituting the known values:

- \( u_y = 5 \, \text{m/s} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( t = 2 \, \text{s} \)

Calculating \( h \):

\[ h = 5 \times 2 - \frac{1}{2} \times 10 \times (2)^2 \]
\[ h = 10 - \frac{1}{2} \times 10 \times 4 \]
\[ h = 10 - 20 = -10 \, \text{m} \]

Since height cannot be negative, we interpret this as the height of the tower being \( 10 \, \text{m} \) above the ground.

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

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