Calculate launch? Please show all the steps & equations?

Question

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. The end of a typical launch ramp is directed 63 degrees above the horizontal. With this launch angle, a skier attains a height of 13.0 m above the end of the ramp. What is the skier's launch speed?

number_______ units_______

Answer 1

The sine of 63 degrees = .8910 therefor the skier covers a distance of D= 13.0 meters/.8910 = 14.6 M to max height.

He would have to be moving fast enough to couter act gravity to the extent of climbing 13 meters

D= 1/2 aT^2 ;  13.0m =1/2(9.8)T^2 = 4.9T^2 ;  T^2= 13.0/4.9

T^2 = 2.653  ;   T= 1.63 second

the skier would have to move a distance of 14.6 meters in 1.63 seconds   14.6/1.63= 8.957 m/s

skiers speed at take-off   8.957 m/s

Answer 2

initial kinetic energy in vertical direction is E=0.5mv^2;

this E turns into potential energy at vertex E=mgh, where h=13m;

thence E=E; 0.5mv^2=mgh,

hence initial component of speed in vertical direction is

v=√(2gh);

therefore launch speed is v0=v/sin(63°) =

=√(2*9.8*13) / sin(63) = 17.92 m/s; - draw a pic!

Answer 3

As the skier leaves the ramp at 63 degrees consider the vertical component of his  velocity.

Use the equation   v^2   =   u^2   +  2*a*s

Since vertical motion stops at max. height, and "a" negative, then

                                   0   =   u^2   -   2 * 9.8 * 13

       u^2   =   254.8    and   u   =   15.96 m/sec.

This is the vertical component of the skier's  velocity (v) as he leaves the ramp

    So    v * sin(63)   =   15.96

       v   =   15.96 / sin(63)   =   17.91 m/sec.