A student throws soft balls out of the window at different angles to the horizontal. All soft balls have the same initial speed `v = 10sqrt3 ms^(-1)`. It turns out that all soft balls landing velocities make angles `30^@` or greater with the horizontal. Find the height h (in m) of the window above the ground.

Explanation

To solve the problem, we need to determine the height \( h \) of the window above the ground from which the soft balls are thrown. Given that the initial speed of the balls is \( v = 10\sqrt{3} \, \text{m/s} \) and that the landing velocities make angles of \( 30^\circ \) or greater with the horizontal, we can analyze the situation step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**:
The balls are thrown from a height \( h \) at different angles, but we need to find the minimum height \( h \) such that the landing angle is \( 30^\circ \) or greater. The worst-case scenario for the landing angle occurs when the ball is thrown horizontally.

2. **Identify the Components of Velocity**:
When the ball is thrown horizontally, the horizontal component of the initial velocity \( u_x = 10\sqrt{3} \, \text{m/s} \) remains constant throughout the motion. The vertical component of the initial velocity \( u_y = 0 \, \text{m/s} \) since it is thrown horizontally.

3. **Determine the Vertical Velocity at Impact**:
After a time \( t \), the vertical velocity \( v_y \) can be calculated using the equation:
\[ v_y = u_y + g t = 0 + g t = g t \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

4. **Using the Angle of Impact**:
The tangent of the angle of impact \( \theta \) is given by:
\[ \tan(\theta) = \frac{v_y}{u_x} \]
For \( \theta = 30^\circ \):
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \implies \frac{v_y}{u_x} = \frac{1}{\sqrt{3}} \]
Substituting the values:
\[ \frac{g t}{10\sqrt{3}} = \frac{1}{\sqrt{3}} \]

5. **Solving for Time \( t \)**:
Rearranging gives:
\[ g t = 10 \implies t = \frac{10}{g} = 1 \, \text{s} \quad (\text{since } g \approx 10 \, \text{m/s}^2) \]

6. **Calculating the Height \( h \)**:
Now, we can find the height \( h \) using the equation of motion for vertical displacement:
\[ h = u_y t + \frac{1}{2} g t^2 \]
Since \( u_y = 0 \):
\[ h = 0 + \frac{1}{2} g t^2 = \frac{1}{2} \times 10 \times (1)^2 = 5 \, \text{m} \]

### Final Answer:
The height \( h \) of the window above the ground is \( 5 \, \text{m} \).

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