The 12 m goes into Potential energy, whilst the horizontal velocity is constant.
1/2 m V^2=mg(12)+1/2 m Vcos66
solve for V.
V^2-Vcos66-24*9.8=0
Use the quadratic equation
The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers speed down a ramp that slopes sharply upward at the end. The sharp upward slope launches them into the air, where they perform acrobatic maneuvers. In the women's competition, the end of a typical launch ramp is directed 66° above the horizontal. With this launch angle, a skier attains a height of 12 m above the end of the ramp. What is the skier's launch speed?
2 answers
the skier's upward velocity is v, so the upward component is v*sin 66° = .9135v
so, if the ramp ends at (0,0), the skier's height is
h = .9135v*t - 4.9t^2
= t(.9135v-4.9t)
roots are at t=0,0.1864v
max h is achieved midway between the roots, at t = 0.0932v
12 = .9135v(.0932v) - 4.9(.0932v)^2
v = 16.7884
so, if the ramp ends at (0,0), the skier's height is
h = .9135v*t - 4.9t^2
= t(.9135v-4.9t)
roots are at t=0,0.1864v
max h is achieved midway between the roots, at t = 0.0932v
12 = .9135v(.0932v) - 4.9(.0932v)^2
v = 16.7884
0 answers