Question

A 60 kg gymnast jumps vertically upward from the edge of a 3.0 m high platform with a speed of 3.5 m/s and lands on a trampoline below.

(a) What is the gymnast's initial total energy?
______ J
(b) How high above the trampoline does he rise?
_______ m
(c) How fast is he going as he lands on the trampoline?
_______ m/s
(d) If the trampoline behaves like a spring with a spring constant of 5.2 104 N/m, how far does the trampoline stretch? (Ignore the small change in gravitational energy when the trampoline is depressed.)
_______ m

Answer 1

(a) E_i = K_i +U_i

= ½ mv_i² + mgh_i

= ½ 60*3.5² + (60*9.8*3)

= 2131.5 J

(b) Remember the kinematic equation vf^2 = vi^2 + 2ad. We know that the ending velocity (when he lands on the trampoline) will be zero, so:

vf^2 = vi^2 + 2ad

0^2 = 3.5^2 + 2 * (-9.8) * d

0 = 12.25 – 19.6d

d = 0.625 m

(c) When he lands, he has 0 PE and all energy is converted to KE:

E_f = E_i (by energy conservation)

=> ½ mv² = 2131.5

=> v = sqrt(2*2131.5/60)

= 8.43 m/s

(d) His KE gets converted into PE stored in the trampoline

=> (1/2)mv^2 = (1/2)kx^2

=> x = v ?(m/k)

=> x = (8.43) * ?[60/(5.2x10^4]

=> x = 0.286 m.