physcis
posted by KayLyn on .
After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 35.5 m horizontally from the end of the ramp. His velocity, just before landing, is 18.5 m/s and points in a direction 43.2 ° below the horizontal. Neglecting air resistance and any lift that he experiences while airborne, find (a) the magnitude and (b) the direction of his initial velocity when he left the end of the ramp.

A = 360  43.2 = 316.8 Deg,CCW.
Vf = 18.5m/s @ 316.8 Deg.
Xf = Xo = 18.5*cos316.8 = 13.49 m/s.
Yf = 18.5*sin316.8 = 12.66 m/s.
Xo * Tf = 35.5 m.
13.49*Tf = 35.5,
Tf = 35.5 / 13.49 = 2.63 s. = Fall time.
Yo + g*Tf = Yf = 12.66 m/s,
Yo + 9.8*2.63 = 12.66,
Yo + 25.8 = 12.66
Yo = 12.66  25.8 = 38.5 m/s.
a. Mag. = sqrt((13.49^2+(38.5)^2 =
40.8 m/s.
b. tanB = Yo/Xo = 38.5 / 13.49 = 2.8510,
B=70.67 South of East=289.3 Deg,CCW.