Physic question below?

Question

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 11 m above the end of the ramp. What is the skier's launch speed?

Answer 1

use energy conservation
sum of all energy is 0
so potential energy at the top of the ramp (mgh) is converted into kinetic energy (.5 m v^2) and then back to potential energy. set KE and PE equal to each other mgh = .5mv2. masses cancel, you are left with gh = .5v^2, where g is acceleration due to gravity (9.8 m/s^2), h is the height (11 meters) , and v is her velocity, or speed.

You really ought to learn how to do these if your future has anything to do with physics, this is pretty basic stuff

Answer 2

EDIT--PLEASE NOTE: My comment about the above response was not correct. I said:
"There is a problem with the analysis above, although correct in principle. The problem stated that the skier reaches 11m above "the end of the ramp", and also that the ramp slopes upward at the end. The potential energy must be measured from the lowest point of travel, which is not the end of the ramp, but somwhere before (which we don't know)." There is a problem with the response, but not that.

You CAN use the conservation of energy method as stated, but you have to take into account that the energy at the peak of the trajectory is not just m*g*h, because the skier still has a horizontal velocity. Therefore the energy at peak altitude is m*g*h + .5*m*(v0*cosT)^2. This can be equated to the kinetic energy at launch, .5*m*v0^2 to find v0.

This problem can also be solved using the angle of launch and altitude given. The veritical component of the launch velocity is v0sinT. Vertically, the velocity will be reduced by gravity so the vertical component of velocity during the trajectory will be v0*sinT-a*t, where t is the time. At the top of the trajectory, the vertical velocity component is zero, so v0*sinT = a*tp, where tp is the time it takes to reach the peak.

The distance traveled vertically is the time integral of the vertical velocity, or v0*t*sinT - .5*a*t^2. Set this equal to the altitude reached (11m) and solve for tp, the time it takes to reach the peak altitude. Then plug this result into the first equation and solve for v0.

EDIT: Hint, there is an easy way to get the solution. Note that if the max height ym = v0*tp*sinT - .5*a*tp^2, you can multiply this equation by a to get a*ym = v0*a*tp*sinT - .5*a^2*tp*2. Then look at the equation at the end of the first paragraph and see that a*tp = v0*sinT. Make the substitution for a*tp and solve for v0.