# A 65 kg tranpoline artist Jumps

• jphillip
In summary, a 65 kg trampoline artist jumps vertically upward from a platform with a speed of 5.0 m/s. Upon landing on the trampoline, he is moving at a speed of 9.16 m/s. The trampoline behaves like a spring with a stiffness of 6.2 x 10^4 N/m. Using conservation of energy, the artist's displacement h below the unstretched position is found to be 0.307 m. This can also be calculated using the artist's speed at trampoline impact and the fully deformed trampoline at its low point.
jphillip

## Homework Statement

65 kg trampoline artist jumps vertically upward from the top of a platform with the speed of 5.0 m/s. a.)How fast is he going as he lands on the trampoline and b.) If the trampoline behaves like a spring with stiffness of 6.2 X 104 N/m how far is it depressed

## Homework Equations

v2 = v2i + 2a(y)
mgh+1/2mv2 = 1/2Km2

## The Attempt at a Solution

a. (5.0m/s)2 + 2(9.8)(3.0) = $\sqrt{83.8}$ => 9.16 m/s2

b. Using conservation of energy I found this on the internet and it is the right answer at the bottom but I can not figure out how they got it.

How did they calculate in the below equation to get 637h + 2723.5

(65)[(9.8)(3.0 + h) + (0.5)(5.0)^2] comes out to 637h + 2723.5

Total energy = mgh + 1/2 mv^2 <=== Note that v=-5.0 m/s
Total energy = (65)[(9.8)(3.0 + h) + (0.5)(5.0)^2] = 1/2 kh^2
NOTE: we have to add h to the potential energy because the trampoline/spring will be compressed by a distance h

(65)[(9.8)(3.0 + h) + (0.5)(5.0)^2] = (0.5)(6.2 x 10^4)h^2
637 h + 2723.5 = (3.1 x10^4)h^2

(3.1 x 10^4) h^2 - 637 h - 2723.5 = 0

Use quadratic formula to solve for h: a= 3.1x 10^4; b= -637; c=-2723.5
h = [637 +/- sqrt{(637)^2 - 4(3.1 x 10^4)(-2723.5)}]/(6.2 x 10^4)
h = [637 +/- sqrt(338119769)]/(6.2 x 10^4)
h = [637 +/- 18388]/(6.2 x 10^4)
h= 0.307 m <=== Amount trampoline is depressed

jphillip said:

## Homework Statement

65 kg trampoline artist jumps vertically upward from the top of a platform with the speed of 5.0 m/s. a.)How fast is he going as he lands on the trampoline and b.) If the trampoline behaves like a spring with stiffness of 6.2 X 104 N/m how far is it depressed

## Homework Equations

v2 = v2i + 2a(y)
mgh+1/2mv2 = 1/2Km2

## The Attempt at a Solution

a. (5.0m/s)2 + 2(9.8)(3.0) = $\sqrt{83.8}$ => 9.16 m/s2
It appears that the trampoline is 3 m below the platform, which you neglected to mention. If this is the case, your calculation for the artist's speed at his impact with the trampoline is correct.
b. Using conservation of energy I found this on the internet and it is the right answer at the bottom but I can not figure out how they got it.

How did they calculate in the below equation to get 637h + 2723.5
it is calculated using basic algebra in the equation for initial energy at the platform in reference to the low point of the trampoline after impact, as shown.
(65)[(9.8)(3.0 + h) + (0.5)(5.0)^2] comes out to 637h + 2723.5

Total energy = mgh + 1/2 mv^2 <=== Note that v=-5.0 m/s
Total energy = (65)[(9.8)(3.0 + h) + (0.5)(5.0)^2] = 1/2 kh^2
NOTE: we have to add h to the potential energy because the trampoline/spring will be compressed by a distance h

(65)[(9.8)(3.0 + h) + (0.5)(5.0)^2] = (0.5)(6.2 x 10^4)h^2
637 h + 2723.5 = (3.1 x10^4)h^2

(3.1 x 10^4) h^2 - 637 h - 2723.5 = 0

Use quadratic formula to solve for h: a= 3.1x 10^4; b= -637; c=-2723.5
h = [637 +/- sqrt{(637)^2 - 4(3.1 x 10^4)(-2723.5)}]/(6.2 x 10^4)
h = [637 +/- sqrt(338119769)]/(6.2 x 10^4)
h = [637 +/- 18388]/(6.2 x 10^4)
h= 0.307 m <=== Amount trampoline is depressed
Since you know the artist's speed at trampoline impact, it is algebraically easier to use conservation of energy between the unstretched trampoline at impact and the fully deformed trampoline at its low point h below the unstretched position (at which point his displacement is x and his speed is ___?).

## 1. How high does a 65 kg trampoline artist jump?

The height of a trampoline jump depends on various factors such as the strength and technique of the artist, as well as the type and quality of the trampoline. However, on average, a 65 kg trampoline artist can jump up to 15 feet high.

## 2. Is it safe for a 65 kg trampoline artist to jump on a trampoline?

Yes, as long as proper safety measures are followed and the trampoline is in good condition, it is safe for a 65 kg trampoline artist to jump on a trampoline. It is important for the artist to warm up and stretch before jumping and to avoid attempting more advanced tricks without proper training and supervision.

## 3. How much force does a 65 kg trampoline artist exert while jumping?

The amount of force exerted by a trampoline artist while jumping varies depending on the height of the jump and the type of landing. However, on average, a 65 kg trampoline artist can exert up to 5 times their body weight while jumping on a trampoline.

## 4. How does a trampoline help a 65 kg trampoline artist in their performance?

A trampoline helps a 65 kg trampoline artist by providing a bouncier surface to perform on and allowing them to jump higher and perform more advanced tricks. It also helps to reduce the impact on their joints, making it easier to perform multiple jumps and stunts without getting injured.

## 5. Can a 65 kg trampoline artist jump in any weather condition?

No, it is not safe for a trampoline artist to jump in extreme weather conditions such as high winds, rain, or snow. These conditions can affect the stability of the trampoline and increase the risk of injury. It is best to only jump on a trampoline in mild weather conditions.

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