Questions: W25 Homework 2 Begin Date: 1/11/2025 2:30:00 PM Due Date: 1/22/2025 11:59:00 PM End Date: 1/22/2025 11:59:00 PM Problem 16: (6% of Assignment Value) A steel ball is dropped onto a hard floor from a height of 1.525 m and rebounds to a height of 1.35 m. Use a coordinate system in which the vertical component of velocity is positive in the upwards direction. - Part (a) Calculate its velocity, in meters per second, just before it strikes the floor. v=-5.47 v=-5.470 Part (b) Calculate its average acceleration, in meters per second squared, during its contact with the floor if that contact lasts 0.0800 ms (8.00 x 10^-5 s). a=

Solution Steps

Using the kinematic equation for free fall: \[ v^2 = u^2 + 2gh \] where \( u = 0 \) (initial velocity), \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity), and \( h = 1.525 \, \text{m} \) (height).

\[ v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1.525} \approx 5.47 \, \text{m/s} \] Since the ball is falling, the velocity just before impact is \( v = -5.47 \, \text{m/s} \).

Using the kinematic equation for upward motion: \[ v^2 = u^2 + 2gh \] where \( u = 0 \) (initial velocity), \( g = -9.8 \, \text{m/s}^2 \) (acceleration due to gravity), and \( h = 1.35 \, \text{m} \) (height).

\[ v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1.35} \approx 5.14 \, \text{m/s} \] Since the ball is moving upwards, the velocity just after impact is \( v = 5.14 \, \text{m/s} \).

Using the formula for average acceleration: \[ a = \frac{\Delta v}{\Delta t} \] where \( \Delta v = v_{\text{final}} - v_{\text{initial}} = 5.14 - (-5.47) = 10.61 \, \text{m/s} \) and \( \Delta t = 8.00 \times 10^{-5} \, \text{s} \).

\[ a = \frac{10.61}{8.00 \times 10^{-5}} \approx 1.33 \times 10^5 \, \text{m/s}^2 \]

Final Answer