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)]`

Explanation

To solve the problem of how far from the throwing point the ball will be at the height of 10 m from the ground, we can follow these steps:

### Step 1: Analyze the problem
The ball is thrown from a height of 10 m with an initial speed of 10 m/s at an angle of 30 degrees. We need to find the horizontal distance traveled by the ball when it reaches the height of 10 m again (which is the same height from which it was thrown).

### Step 2: Break down the initial velocity into components
The initial velocity \( u \) can be broken down into horizontal and vertical components:
- Horizontal component \( u_x = u \cdot \cos(\theta) = 10 \cdot \cos(30^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \)
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