Before beginning this problem, review Interactive Solution 3.45. After leaving t

Question

Before beginning this problem, review Interactive Solution 3.45. After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 53.7 m horizontally from the end of the ramp. His velocity, just before landing, is 19.2 m/s and points in a direction 32.4 degree 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) Number Units (b) Number Units

Explanation / Answer

horizontal velocity will not change during the flight as air resistance is not important.

when he lands his horizontal velocity is

19.2*cos(32.4) = 16.2 m/s

the time in flight comes from

x = Vht

53.7 = 16.2t

t = 3.315 seconds

his vertical velocity at landing is

19.2sin(32.4) = -10.288 m/s

his vertical velocity equation is

v = V0 - gt

-10.288 = V0 - 9.81(3.315)

V0 = -10.288 + 9.81(3.315)

V0 = 22.23 m/s

his velocity magnitude is

v = sqrt(22.23^2 + 16.2^2)

v = 27.5 m/s    ................Ans(1)

his initial direction was

tan = 16.2/22.23

= 36.1 degrees above the horizontal    ...........Ans(2)

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