V = 22m/s[-40.3o]
Xo = 22*cos(-40.3) = 16.78 m/s. = Hor.
component of initial velocity.
Y = 22*sin(-40.3) = -14.23 m/s.
Xo * Tf = 62.3 m.
16.78 * Tf = 62.3
Tf = 3.71 s. = Fall time.
Y = Yo + g*Tf = -14.23 m/s.
a. Yo + 9.8*3.71 = -14.23
Yo + 36.38 = -14.23
Yo = -14.23 - 36.38 = -50.61 m/s. = Vertical component of initial velocity.
Mag. = Sqrt(Xo^2+Yo^2)
Mag. = Sqrt(16.78^2+50.61^2) = 53.32m/s
b. tanA = Yo/Xo = -50.61/16.78 =-3.016091
A = -71.7o = = 71.7o S. of E = Direction
After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 62.3 m horizontally from the end of the ramp. His velocity, just before landing, is 22.0 m/s and points in a direction 40.3 ° 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.
2 answers
henry your answer is worthless and I hope you've since died in a horrible fire