- #1

jphillip

- 10

- 0

## 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 10

^{4}N/m how far is it depressed

## Homework Equations

v

^{2}= v

^{2}

_{i}+ 2a(y)

mgh+

^{1}/

_{2}mv

^{2}=

^{1}/

_{2}Km

^{2}

## The Attempt at a Solution

a. (5.0m/s)

^{2}+ 2(9.8)(3.0) = [itex]\sqrt{83.8}[/itex] => 9.16 m/s

^{2}

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