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

Explanation

To solve the problem step by step, we need to analyze the motion of the ball thrown from the roof of a building. Here’s how we can do it:

### Step 1: Identify the initial conditions
- The height of the building (h) = 10 m
- The initial speed of the ball (u) = 10 m/s
- The angle of projection (θ) = 30°
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Resolve the initial velocity into components
The initial velocity can be broken down into horizontal and vertical components:
- Horizontal component (u_x) = u * cos(θ) = 10 * cos(30°) = 10 * (√3/2) = 5√3 m/s
- Vertical component (u_y) = u * sin(θ) = 10 * sin(30°) = 10 * (1/2) = 5 m/s

### Step 3: Determine the time of flight until the ball reaches the height of 10 m again
Since the ball is thrown from a height of 10 m and we want to find the time when it returns to the same height, we can use the vertical motion equation:
- The equation of motion in the vertical direction is given by:
\[ y = u_y t - \frac{1}{2} g t^2 \]
- Here, y = 0 (since it returns to the same height), u_y = 5 m/s, and g = 10 m/s².

Setting up the equation:
\[ 0 = 5t - \frac{1}{2} (10) t^2 \]
\[ 0 = 5t - 5t^2 \]
Factoring out t:
\[ t(5 - 5t) = 0 \]
This gives us two solutions:
1. t = 0 (initial time)
2. t = 1 s (time when it returns to the height of 10 m)

### Step 4: Calculate the horizontal distance traveled
Now, we can find the horizontal distance (d) traveled by the ball when it reaches the height of 10 m again:
- The horizontal distance is given by:
\[ d = u_x * t \]
Substituting the values:
\[ d = 5√3 * 1 = 5√3 \text{ m} \]

### Step 5: Calculate the numerical value
Now, we can calculate the numerical value of \( 5√3 \):
- \( √3 \approx 1.732 \)
Thus,
\[ d \approx 5 * 1.732 \approx 8.66 \text{ m} \]

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

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