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 ` 30 ^@ ` with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground?
` (g = 10 m//s ^ 2 ) `

Options

A. (a)8.66 m
B. (b) 5.20 m
C. (c)4.33 m
D. (d) 2.60 m

Explanation

To solve the problem step by step, we will analyze the motion of the ball thrown by the boy from the roof of a 10 m high building.

### Step 1: Understand the problem
The boy throws a ball with an initial speed of 10 m/s at an angle of 30 degrees with the horizontal from a height of 10 m. We need to find how far from the throwing point the ball will be when it reaches the height of 10 m from the ground.

### Step 2: Break down the initial velocity into components
The initial velocity (u) can be broken down into horizontal (u_x) and vertical (u_y) components:
- \( u_x = u \cdot \cos(\theta) \)
- \( u_y = u \cdot \sin(\theta) \)

Given:
- \( u = 10 \, \text{m/s} \)
- \( \theta = 30^\circ \)

Calculating the components:
- \( u_x = 10 \cdot \cos(30^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \)
- \( u_y = 10 \cdot \sin(30^\circ) = 10 \cdot \frac{1}{2} = 5 \, \text{m/s} \)

### Step 3: Determine the time of flight until the ball reaches the height of 10 m
Since the ball is thrown from a height of 10 m and we want to find the distance when it is still at that height, we can use the vertical motion equation:
\[ h = h_0 + u_y t - \frac{1}{2} g t^2 \]

Where:
- \( h = 10 \, \text{m} \) (final height)
- \( h_0 = 10 \, \text{m} \) (initial height)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t \) = time in seconds

Substituting the values:
\[ 10 = 10 + 5t - \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ 0 = 5t - 5t^2 \]
Factoring out:
\[ 0 = 5t(1 - t) \]

This gives us two solutions:
1. \( t = 0 \) (at the moment of throwing)
2. \( t = 1 \) (when it returns to the height of 10 m)

### Step 4: Calculate the horizontal distance traveled
Now, we can find the horizontal distance (R) traveled by the ball when it reaches the height of 10 m again, using the horizontal component of the velocity:
\[ R = u_x \cdot t \]

Substituting the values:
- \( u_x = 5\sqrt{3} \, \text{m/s} \)
- \( t = 1 \, \text{s} \)

Calculating the distance:
\[ R = 5\sqrt{3} \cdot 1 = 5\sqrt{3} \, \text{m} \]

### Step 5: Numerical approximation
To find the numerical value:
\[ R \approx 5 \cdot 1.732 = 8.66 \, \text{m} \]

### Final Answer
The ball will be approximately **8.66 m** away from the throwing point when it reaches the height of 10 m from the ground.

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