A 65 kg tranpoline artist Jumps

[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 = v²_i + 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

Rewriting into standard quadratic form:
(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

 

[PhanthomJay]

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 = v²_i + 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

Rewriting into standard quadratic form:
(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 ___?).