A ball is thrown from the top of a tower with an intial velocity of 10 m//s at an angle `37^(@)` above the horizontal, hits the ground at a distance 16 m from the base of tower. Calculate height of tower. `[g=10 m//s^(2)]`

Explanation

To solve the problem of finding the height of the tower from which a ball is thrown, we can break down the motion into horizontal and vertical components. Here’s a step-by-step solution:

### Step 1: Resolve the initial velocity into horizontal and vertical components.
The ball is thrown with an initial velocity \( u = 10 \, \text{m/s} \) at an angle of \( 37^\circ \) above the horizontal.

- **Horizontal component** \( u_x = u \cos(37^\circ) \)
- **Vertical component** \( u_y = u \sin(37^\circ) \)

Using the values:
- \( \cos(37^\circ) = \frac{4}{5} \)
- \( \sin(37^\circ) = \frac{3}{5} \)

Calculating the components:
\[ u_x = 10 \cdot \frac{4}{5} = 8 \, \text{m/s} \]
\[ u_y = 10 \cdot \frac{3}{5} = 6 \, \text{m/s} \]

### Step 2: Determine the time of flight.
The horizontal distance covered by the ball is given as \( d = 16 \, \text{m} \). The time \( t \) taken to cover this distance can be calculated using the formula:
\[ d = u_x \cdot t \]
Rearranging gives:
\[ t = \frac{d}{u_x} = \frac{16}{8} = 2 \, \text{s} \]

### Step 3: Calculate the height of the tower.
Using the vertical motion equation to find the height \( h \):
\[ h = u_y t - \frac{1}{2} g t^2 \]
Substituting the known values:
- \( u_y = 6 \, \text{m/s} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( t = 2 \, \text{s} \)

Calculating:
\[ h = 6 \cdot 2 - \frac{1}{2} \cdot 10 \cdot (2^2) \]
\[ h = 12 - \frac{1}{2} \cdot 10 \cdot 4 \]
\[ h = 12 - 20 \]
\[ h = -8 \, \text{m} \]

Since height cannot be negative, we interpret this as the height of the tower being \( 8 \, \text{m} \) (the negative sign indicates that the displacement is downward).

### Final Answer:
The height of the tower is \( 8 \, \text{m} \).

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