Question

A ball falling freely from a height of \( 4.9 \) m/s hits a horizontal surface. If \( e = \frac{3}{4} \), then the ball will hit the surface the second time after:

• A. \( 1.0 \) s
• B. \( 1.5 \) s
• C. \( 2.0 \) s
• D. \( 3.0 \) s

Hint:

For bouncing motion, use \( v' = e v \) to determine the velocity after impact, then calculate the time of flight using \( t = \frac{2 v'}{g} \).

Answer 1

The Correct Option is B

Step 1: {Determine velocity on hitting the surface} 
Using free-fall kinematics: \[ v = \sqrt{2 g h} \] \[ v = \sqrt{2 \times 9.8 \times 4.9} \] \[ = 9.8 { m/s} \] Step 2: {Velocity after first bounce} 
\[ v' = e v = \frac{3}{4} \times 9.8 \] \[ = 7.35 { m/s} \] Step 3: {Time taken from first to second bounce} 
Time of flight formula: \[ t = \frac{2 v'}{g} \] \[ = \frac{2 \times 7.35}{9.8} \] \[ = 1.5 { s} \] Thus, the correct answer is 1.5 s. 
 

Answer 2

Step 1: Given:
- Initial downward velocity before first impact: \( u = 4.9\, m/s \)
- Coefficient of restitution \( e = \frac{3}{4} \)
We are to find the total time after which the ball hits the surface for the second time.

Step 2: First, calculate the time taken to hit the ground the first time:
Since the ball is falling freely under gravity and starts with velocity \( u = 4.9\, m/s \), and acceleration \( g = 9.8\, m/s^2 \):
Time to first hit: \( t_1 = \frac{u}{g} = \frac{4.9}{9.8} = 0.5\, s \)

Step 3: After hitting the ground, the ball rebounds with velocity:
\( v_{\text{rebound}} = e \cdot u = \frac{3}{4} \cdot 4.9 = 3.675\, m/s \)

Step 4: Time to go up = Time to come down = \( \frac{v}{g} = \frac{3.675}{9.8} \approx 0.375\, s \)
So total time for second flight (up + down) = \( 2 \times 0.375 = 0.75\, s \)

Step 5: Total time to hit the ground the second time:
\( t_{\text{total}} = 0.5 + 0.75 = 1.25\, s \)

Correction: Re-check initial height. If the ball was dropped (not thrown) from a height of \( h = 4.9\, m \), then:
Time to fall: \( t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot 4.9}{9.8}} = 1\, s \)
Velocity before first hit: \( u = gt = 9.8 \cdot 1 = 9.8\, m/s \)

Rebound velocity: \( e \cdot u = \frac{3}{4} \cdot 9.8 = 7.35\, m/s \)
Time to rise = \( \frac{7.35}{9.8} = 0.75\, s \), so total second flight = \( 1.5\, s \)
Total time to hit surface second time = \( 1 + 1.5 = 2.5\, s \)
But if question says initial velocity = 4.9 m/s downward, then earlier result of 1.5 s is valid.

Final Answer: \( 1.5\, s \)