A boy playing on the roof of a `10 m` high building throws a ball with a speed of `10 m//s` at an angle of `30(@)` with the horizontal. How far from the throwing point will the ball be at the height of `10 m` from the ground ?
`[ g = 10m//s^(2) , sin 30^(@) = (1)/(2) , cos 30^(@) = (sqrt(3))/(2)]`

Options

A. ` 5.20 m`
B. `4.33 m`
C. ` 2.60 m`
D. ` 8.66 m`

Explanation

To solve the problem step by step, we need to analyze the motion of the ball thrown by the boy from the roof of the building. We will break down the motion into horizontal and vertical components and use the kinematic equations to find the horizontal distance traveled by the ball when it reaches the height of 10 m from the ground.

### Step 1: Determine the initial velocity components
The initial velocity \( u \) of the ball is given as \( 10 \, \text{m/s} \) and the angle of projection \( \theta \) is \( 30^\circ \).

- The horizontal component of the initial velocity \( u_x \) is given by:
\[ u_x = u \cdot \cos(\theta) = 10 \cdot \cos(30^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \]

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

### Step 2: Determine the time of flight to reach the height of 10 m
Since the ball is thrown from a height of 10 m and we want to find the time it takes to reach the same height again, we can use the kinematic equation for vertical motion:
\[ h = u_y t - \frac{1}{2} g t^2 \]
Here, \( h = 0 \) (the height from which it was thrown), \( u_y = 5 \, \text{m/s} \), \( g = 10 \, \text{m/s}^2 \).

Setting up the equation:
\[ 0 = 5t - \frac{1}{2} \cdot 10 t^2 \]
This simplifies to:
\[ 0 = 5t - 5t^2 \]
Factoring out \( 5t \):
\[ 5t(1 - t) = 0 \]
Thus, \( t = 0 \) (initial time) or \( t = 1 \, \text{s} \) (time to reach the height of 10 m again).

### Step 3: Calculate the horizontal distance traveled
Now that we have the time \( t = 1 \, \text{s} \), we can calculate the horizontal distance \( d \) traveled using the horizontal component of the velocity:
\[ d = u_x \cdot t = 5\sqrt{3} \cdot 1 = 5\sqrt{3} \, \text{m} \]
Using the approximation \( \sqrt{3} \approx 1.732 \):
\[ d \approx 5 \cdot 1.732 \approx 8.66 \, \text{m} \]

### Conclusion
The ball will be approximately \( 8.66 \, \text{m} \) away from the throwing point when it reaches the height of \( 10 \, \text{m} \) from the ground.

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