Question

From a height of  'h' above the ground, a ball is projected up at an angle \( 30^\circ \) with the horizontal. If the ball strikes the ground with a speed of 1.25 times its initial speed of \( 40 \ ms^{-1} \), the value of 'h' is:

• A. \( 75 \ m \)
• B. \( 60 \ m \)
• C. \( 30 \ m \)
• D. \( 45 \ m \)

Hint:

For projectile motion problems involving energy conservation, use the equation \( \frac{1}{2} m u^2 + mg h = \frac{1}{2} m v^2 \) to relate initial and final energy states.

Answer 1

The Correct Option is D

Step 1: Given Data 
- Initial speed of the ball: \( u = 40 \ ms^{-1} \) - Projection angle: \( \theta = 30^\circ \) - Final speed at impact: \( v = 1.25 \times 40 = 50 \ ms^{-1} \) - Acceleration due to gravity: \( g = 10 \ ms^{-2} \) - Height from which the ball is projected: \( h \) 

Step 2: Using the Energy Conservation Principle 
The total mechanical energy at the initial and final points must be equal: \[ \frac{1}{2} m u^2 + mg h = \frac{1}{2} m v^2. \] Canceling \( m \) on both sides: \[ \frac{1}{2} u^2 + g h = \frac{1}{2} v^2. \] Substituting the given values: \[ \frac{1}{2} (40)^2 + 10h = \frac{1}{2} (50)^2. \] 

Step 3: Solving for \( h \) 
\[ \frac{1}{2} (1600) + 10h = \frac{1}{2} (2500). \] \[ 800 + 10h = 1250. \] \[ 10h = 1250 - 800. \] \[ 10h = 450. \] \[ h = 45 \ m. \] 

Step 4: Conclusion 
Thus, the height \( h \) from which the ball is projected is: \[ \mathbf{45 \ m}. \]