Question

A 65 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5 m/s. how fast is he going as he lands on the trampoline 3 m below if the trampoline behaves like a spring with spring stiffness constant 6.2 x 104 n/m, how far does he depress it?

Answer 1

m = mass of trampoline artist = 65 kg

v₀ = initial speed of the artist at the top of platform = 5 $\frac{m }{s}$

h = height through which the artist drop before landing on trampoline = 3 m

v = final speed of the artist just before landing on trampoline = ?

using conservation of energy

Kinetic energy of artist just before landing = initial kinetic energy at the top of platform + potential energy at the top of platform

(0.5) m v² = (0.5) m v₀² + mgh

dividing each term by "m"

(0.5)  v² = (0.5) v₀² + gh

inserting the values

(0.5)  v² = (0.5) (5)² + (9.8)(3)

v = 9.2 $\frac{m }{s}$

x = compression of the trampoline = ?

k = spring stiffness constant = 62000 $\frac{N }{m}$

Assuming the lowest depression point as reference line for measuring the potential energy

using conservation of energy

kinetic energy of artist + potential energy of artist before landing = spring potential energy of trampoline

(0.5) m v² + mg x = (0.5) k x²

inserting the values

(0.5) (65) (9.2)² + (65 x 9.8) x = (0.5) (62000) x²

x = 0.31 m