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: Calculate the velocity upon surface impact. Employing free-fall kinematics yields: \[ v = \sqrt{2 g h} \] \[ v = \sqrt{2 \times 9.8 \times 4.9} \] \[ = 9.8 { m/s} \] Step 2: Determine the velocity post-first bounce. The rebound velocity is given by: \[ v' = e v = \frac{3}{4} \times 9.8 \] \[ = 7.35 { m/s} \] Step 3: Calculate the time interval between the first and second bounce. Using the time of flight formula: \[ t = \frac{2 v'}{g} \] \[ = \frac{2 \times 7.35}{9.8} \] \[ = 1.5 { s} \] Consequently, the final answer is 1.5 s.