A 65.0-kg bungee jumper steps off a bridge with a light bungee cord tied to hers

Question

A 65.0-kg bungee jumper steps off a bridge with a light bungee cord tied to herself and the bridge. The unstretched length of the cord is 11.0m. She reaches the bottom of her motion 36.0m below the bridge before bouncing back up. (Her descent can be separated into 11.0m free fall and 25.0m Simple Harmonic Motion). For the time of stretching let y(t)=Acos(omegat+phi) where the positive direction is downward, the equilibrium position is y=0, and the time t=0 at the moment the cord begins to stretched.

a. Find (spring constant) k
b. How far below the bridge is the location of the equilibrium point where spring and gravitational force are balanced.
c. angular frequency (omega)
d. For SHM (when the cord stretches) find amplitude A
e. phase shift (phi)
f. Total time of descent.

i think i have parts a,b,c and f but i cant get my head around d or e.
The values i currently have are
a. 22.4 N/m
b. 28.4m
c. 0.59 rad/s
f. 12.1s

Explanation / Answer

(a) t = ?(2h/g) = ?(22m / 9.8m/s

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