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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.
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.
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