A 60 kg gymnast jumps vertically upward from the edge of a 3.0 m high platform w
ID: 2032868 • Letter: A
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
Explanation / Answer
(a) Ei = Ki +Ui
= ½ mvi2 + mghi
= 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:
Ef = Ei (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.
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