After leaving the end of a ski ramp, a ski jumper landsdownhill at a point that

Question

After leaving the end of a ski ramp, a ski jumper landsdownhill at a point that is displaced 55km horizontally from theend of the ramp. His velocity, just before landing, is 25m/sand points in a direction 38o below thehorizontal. Neglecting air resistance and any lift that heexperiences while airborne, find his initial velocity (magnitudeand direction) when he left the end of the ramp. answer: 23m/s, 31oabove the horizontal After leaving the end of a ski ramp, a ski jumper landsdownhill at a point that is displaced 55km horizontally from theend of the ramp. His velocity, just before landing, is 25m/sand points in a direction 38o below thehorizontal. Neglecting air resistance and any lift that heexperiences while airborne, find his initial velocity (magnitudeand direction) when he left the end of the ramp. answer: 23m/s, 31oabove the horizontal

Explanation / Answer

    Final velocity, V = 25 m/s     Angle of reaching the ground, = 38 deg (below horizontal )     Horizontal distance, X = 55 m     Let     Initial velocity of projection = U     Angle of projection =     Horizontal component of final velocity, Vx = Vcos = 25 cos 38 = 19.7 m/s     Vertical component of final velocity, Vy = - Vsin = - 25 sin 38 = - 15.4 m/s     Horizontal component of initial velocity, Ux =Vx = 19.7 m/s ...................(1)     Time taken, t = X / Ux = 55 / 19.7 = 2.79 s     Vy = Uy + at     - 15.4 = Uy - gt     - 15.4 = Uy - 9.8 * 2.79     Uy = 11.942m/s         ........................(2)     Initial velocity of projection, U = [ (Ux ) 2 + ( Uy ) 2 ]                                                   = 23 m/s         Angle of projection, = Arc tan ( Uy / Ux) = 31o ( above horizontal )

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