The brothers of Iota Eta Pi fraternity build a platform, supported at all four corners by vertical springs, in the basement of their frat house. A brave fraternity brother wearing a football helmet stands in the middle of the platform; his weight compresses the springs by a distance 0.16m . Then four of his fraternity brothers, pushing down at the corners of the platform, compress the springs another distance 0.57m until the top of the brave brother's helmet is a distance 0.80m below the basement ceiling. They then simultaneously release the platform. You can ignore the masses of the springs and platform.

A) When the dust clears, the fraternity asks you to calculate their fraternity brother's speed just before his helmet hit the flimsy ceiling.
B) Without the ceiling, how high would he have gone?
C) In discussing their probation, the dean of students suggests that the next time they try this, they do it outdoors on another planet. Would the answer to part (B) be the same if this stunt were performed on a planet with a different value of g ? Assume that the fraternity brothers push the platform down 0.57m as before. (yes/no)
D) Explain your reasoning.

The first distance given allows you to calculate the spring constant of the construction. (it will be in terms of the brother's mass; don't worry, it factors out later.)

k = m×g / x where x = 0.16m

When the rest of the brothers pull the platform lower, you can calculate the stored energy.

Us = 0.5× k×X²

Us = 0.5×m×g× x where x = 0.16m

When the brothers release the platform, the stored energy is converted to kinetic energy and gravitational potential energy of the brave fraternity brother (BFB) until the springs reach equilibrium position.

PEgrav = mgh where h = 0.16m

KE = 0.5×mv²

At equilibrium position, the energy originally stored in the spring will be equal to the energy of the BFB. Solve for v.

0.5×m×(-9.81 m/s²) ×0.16 m = m ×(-9.81 m/s²)×0.16 m + 0.5×m×v²

As promised, mass factors out.

0.5× (-9.81 m/s²) ×0.16m = (-9.81 m/s²) × 0.16 m + 0.5× v²

-0.5 * (-9.81 m/s²) ×0.16m = 0.5 ×V²

9.81 m/s²×0.16m = v²

v = √(6.9651 m/s²)

v = 1.235m/s

A little arithmetic will show that from the equilibrium point, the BFB has 0.80m between his helmet and the ceiling. Calculate his speed after he has traveled this distance.

V = V0 + At

x = x0 + v0t + 0.5At²

Solving for t:

0.19 m = 0 m + 1.235 m/s×t + 0.5×(-9.81m/s²)×t²

You'll find that t = 0.0923 s and 0.378s. The smaller value is the ceiling crossing on the way up, and the larger is on the way back down, assuming no ceiling is actually present.

1. With t, solve for v:

v = 1.235 m/s + (-9.81 m/s²) ×0.0923s

v = 0.3296m/s

2. with initial v, solve for x:

v = v0 + At

t = (0 - 1.235m/s) / -9.81 m/s²

t = 0.126 seconds

x = x0 + v0t + 0.5×At²

x = 0 + (1.235m/s×0.126 s) + (0.5 × (-9.81 m/s²×0.126²s)

x = 0 + 0.15561 m - . 0.0778m

x = 0.077m (above the platform)