Consider a frictionless ski jump. The platform is at a height of h_1=20.0 m off
ID: 1491777 • Letter: C
Question
Consider a frictionless ski jump. The platform is at a height of h_1=20.0 m off the ground, and extends for a length L = 40.0 m. A small ramp is placed at x - L at an angle of theta=10.0degree from the horizontal. The height of the ramp is h_2=5.0m. The initial speed of the ski jumper when she is on the platform is v_i=10.0 m/s. What is the ski jumper's speed immediately after she exits the ramp (v,)? What is the skier's maximum height H after leaving the ramp? What distance does the skier travel in the horizontal direction between when she leaves the ramp and when she touches down on the ground? What is her speed when she reaches the ground (vf)?Explanation / Answer
A) By energy conservation,
Initial total energy = Final T.E.
0.5 m *10^2 + m *9.8*20 = 0.5*m*v^2 +m*9.8*5
cancelling m,
50 + 196 = 0.5v^2 +49
0.5 v^2 = 197
v = sqrt(2*197) = 19.85 m/s
b) at maximum height, speed v is zero,
0.5 m *10^2 + m *9.8*20 = m *9.8* h + 0
50 +196 = 9.8 h
h =246/9.8 = 25.1 m
c) Vvertical= 19.85 sin 10 degree =3.447 m/s
s = ut - 0.5*gt^2
-5 = 3.447 t - 4.9t^2
4.9t^2 - 3.447 t -5 = 0
t = 1.01 s
horizontal distance = v cos 10 degree t =19.85 * 1.01* cos 10 degree
=19.74 m
d) By energy conservation,
Initial total energy = Final T.E. when hits the ground
0.5 m *10^2 + m *9.8*20 = 0.5*m*v^2 + 0
cancelling m,
50 + 196 = 0.5v^2
0.5 v^2 = 246
v = sqrt(2*246) = 22.18 m/s
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