Question

A 61-kg trampoline artist jumps upward from the top of a platform with a vertical speed of 5.1 m/s.

Part A

How fast is he going as he lands on the trampoline, 2 m below?(Figure 1)

Part B

If the trampoline behaves like a spring of spring constant 7.2×10⁴ N/m, how far does he depress it?

Answer 1

Gravitational acceleration = g = 9.81 m/s2

Mass of the artist = m = 61 kg

Initial speed of the artist = V1 = 5.1 m/s

Speed of the artist when he lands on the trampoline = V2

Height of the platform from the trampoline = H = 2 m

By conservation of energy the potential plus kinetic energy of the artist at the platform is equal to the kinetic energy of the artist as he lands on the trampoline.

Spring constant of the trampoline = k = 7.2 x 10^4 N/m

Distance the trampoline depresses = X

By conservation of energy the kinetic energy of the artist as he lands on the trampoline plus the potential energy lost by the artist as the trampoline depresses is equal to the potential energy gained by the trampoline.

X = 0.244 m or -0.227 m

Distance cannot be negative.

X = 0.244 m

A) Speed of the artist as he lands on the trampoline = 8.08 m/s

B) Distance the trampoline depresses due to the artist = 0.244 m