Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

46. Velocity and Distance in Free Fall with Rebound

An object is dropped from a height of 50 meters, falls freely, and rebounds with half the speed with which it hits the ground. Calculate the maximum height it reaches after the rebound.

Sagot :

To calculate the maximum height an object reaches after rebounding when it is dropped from a height of 50 meters and falls freely, we need to follow a systematic approach. Let's break it down step by step:

1. Determine the velocity when the object hits the ground:
- The object starts from rest (initial velocity \( u = 0 \)) and falls freely under the influence of gravity.
- The height from which it falls is 50 meters.
- The acceleration due to gravity (\( g \)) is \( 9.8 \, \text{m/s}^2 \).

To find the velocity (\( v \)) when the object hits the ground, we use the kinematic equation:
[tex]\[ v^2 = u^2 + 2gh \][/tex]
Since the initial velocity \( u = 0 \), the equation simplifies to:
[tex]\[ v^2 = 2gh \][/tex]
Plugging in the values:
[tex]\[ v^2 = 2 \times 9.8 \times 50 \][/tex]
[tex]\[ v = \sqrt{2 \times 9.8 \times 50} \][/tex]
[tex]\[ v \approx 31.30 \, \text{m/s} \][/tex]

2. Calculate the rebound velocity:
- The rebound velocity is stated to be half the speed with which the object hits the ground.
- So, the rebound velocity (\( v_{\text{rebound}} \)) is:
[tex]\[ v_{\text{rebound}} = \frac{v}{2} \][/tex]
[tex]\[ v_{\text{rebound}} = \frac{31.30}{2} \][/tex]
[tex]\[ v_{\text{rebound}} \approx 15.65 \, \text{m/s} \][/tex]

3. Determine the maximum height after the rebound:
- To find the maximum height (\( h_{\text{rebound}} \)) reached after the rebound, we again use the kinematic equations.
- At the maximum height, the final velocity \( v = 0 \), and the initial velocity \( u \) is the rebound velocity.
- We use the equation:
[tex]\[ v^2 = u^2 + 2gh \][/tex]
Here, \( v = 0 \) (at the maximum height) and initial velocity \( u = v_{\text{rebound}} \):

Rearranging the equation to solve for \( h \):
[tex]\[ 0 = v_{\text{rebound}}^2 - 2gh \][/tex]
[tex]\[ h = \frac{v_{\text{rebound}}^2}{2g} \][/tex]

Plugging in the values:
[tex]\[ h = \frac{(15.65)^2}{2 \times 9.8} \][/tex]
[tex]\[ h = \frac{245.10}{19.6} \][/tex]
[tex]\[ h \approx 12.5 \, \text{meters} \][/tex]

Therefore, the maximum height the object reaches after the rebound is approximately 12.5 meters.