Homework Help: physcis

Posted by KayLyn on Sunday, February 19, 2012 at 1:50pm.

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.

physcis - Henry, Monday, February 20, 2012 at 6:31pm
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.

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