Question

A bowling machine placed at a height \( h \) above the earth surface releases different balls with different angles but with the same velocity \( 10 \sqrt{3} \, \text{ms}^{-1} \). All these balls landing velocities make angles 30° or more with horizontal. Then the height \( h \) (in meters) is:

• A. 15
• B. 12
• C. 10
• D. 5

Hint:

For projectile motion problems, break the motion into horizontal and vertical components, and use the standard equations of motion to solve for the unknowns.

Answer 1

The Correct Option is D

To determine the height \( h \) from which the bowling machine releases the balls, we analyze the projectile motion and the given conditions. Given: 
- Initial velocity \( v_0 = 10\sqrt{3} \, \text{ms}^{-1} \) 
- All landing velocities make angles of 30° or more with the horizontal. 

Step 1: Determine the Landing Velocity Components When a projectile lands, its velocity can be broken into horizontal (\( v_x \)) and vertical (\( v_y \)) components. The angle \( \theta \) that the landing velocity makes with the horizontal is given by: \[ \tan \theta = \frac{v_y}{v_x} \] Given that \( \theta \geq 30^\circ \), we have: \[ \tan 30^\circ \leq \frac{v_y}{v_x} \] \[ \frac{1}{\sqrt{3}} \leq \frac{v_y}{v_x} \] 

Step 2: Express \( v_x \) and \( v_y \) in Terms of Initial Conditions The horizontal component of velocity remains constant: \[ v_x = v_0 \cos \alpha \] The vertical component of velocity at landing is: \[ v_y = v_0 \sin \alpha - gt \] where \( \alpha \) is the launch angle and \( t \) is the time of flight. 

Step 3: Time of Flight The time of flight \( t \) for a projectile launched from height \( h \) is given by: \[ h = v_0 \sin \alpha \cdot t - \frac{1}{2} g t^2 \] Solving for \( t \) involves solving a quadratic equation, but we can use the fact that the minimum angle is 30° to find the minimum height. 

Step 4: Minimum Height Calculation For the minimum angle \( \theta = 30^\circ \): \[ \tan 30^\circ = \frac{v_y}{v_x} = \frac{1}{\sqrt{3}} \] Using the relationship between \( v_y \) and \( v_x \): \[ v_y = v_x \cdot \frac{1}{\sqrt{3}} \] Since \( v_x = v_0 \cos \alpha \) and \( v_y = v_0 \sin \alpha - gt \), we can equate: \[ v_0 \sin \alpha - gt = v_0 \cos \alpha \cdot \frac{1}{\sqrt{3}} \] Solving for \( t \) and substituting back into the height equation, we find: \[ h = \frac{v_0^2 \sin^2 \alpha}{2g} \] For the minimum angle \( \alpha = 30^\circ \): \[ h = \frac{(10\sqrt{3})^2 \sin^2 30^\circ}{2 \cdot 9.8} = \frac{300 \cdot 0.25}{19.6} = \frac{75}{19.6} \approx 3.826 \, \text{m} \] However, this does not match the given options. Let's consider the maximum height for the given conditions. 

Step 5: Re-evaluate the Approach Given the complexity, let's use the fact that the minimum height corresponds to the minimum angle. For \( \theta = 30^\circ \): \[ h = \frac{v_0^2 \sin^2 \alpha}{2g} \] With \( \alpha = 30^\circ \): \[ h = \frac{(10\sqrt{3})^2 \cdot 0.25}{2 \cdot 9.8} = \frac{300 \cdot 0.25}{19.6} = \frac{75}{19.6} \approx 3.826 \, \text{m} \] This still does not match the options. Let's consider the maximum height for the given conditions. 

Step 6: Correct Calculation Given the discrepancy, let's use the energy approach. The total energy at launch equals the total energy at landing: \[ \frac{1}{2} m v_0^2 + m g h = \frac{1}{2} m v^2 \] Given \( v = \frac{v_0}{\cos \theta} \) and \( \theta \geq 30^\circ \): \[ \frac{1}{2} m v_0^2 + m g h = \frac{1}{2} m \left( \frac{v_0}{\cos 30^\circ} \right)^2 \] Solving for \( h \): \[ h = \frac{v_0^2}{2g} \left( \frac{1}{\cos^2 30^\circ} - 1 \right) = \frac{(10\sqrt{3})^2}{2 \cdot 9.8} \left( \frac{1}{0.75} - 1 \right) = \frac{300}{19.6} \left( \frac{4}{3} - 1 \right) = \frac{300}{19.6} \cdot \frac{1}{3} \approx 5.1 \, \text{m} \] 

This is closest to option (4). Final Answer: \[ \boxed{5} \]