Question

After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 30 m horizontally from the end of the ramp. his velocity, just before landing, is 19 m/s and points in a direction 43° below the horizontal. neglecting air resistance and any lift he experiences while airborne, find his initial velocity (magnitude and direction), when the he left the end of the ramp.

Answer 1

Answer:

  Initial velocity = 16.15 m/s, 30.63° above horizontal

Explanation:

 We have equation of motion , $s= ut+\frac{1}{2} at^2$, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  The horizontal component of a projectile is constant throughout its journey. If u is the initial velocity, the horizontal velocity = u cos θ is a constant.

  Final horizontal velocity = 19 cos 43 = 13.90 m/s

  So initial horizontal velocity = u cos θ = 13.90 m/s.

  We also have horizontal displacement = 30 meter.

   Substituting in the equation of motion

        $30 = 13.90*t+\frac{1}{2} *0*t^2\\ \\ t=2.16 seconds$

  We also have final vertical velocity = 19*sin 43 = 12.96 m/s, but in vertical direction acceleration due to gravity is there.

  We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

 In case of vertical motion, final velocity = 12.96 m/s, time taken = 2.16 seconds, acceleration = 9.81 $m/s^2$ and initial velocity = u sin θ

      12.96 = u sin θ + 9.81*2.16

      u sin θ = -8.23 m/s = 8.23 m/s (upward direction)

    So magnitude of initial velocity = $\sqrt{13.90^2+8.23^2} =16.15m/s$

     Direction, θ = tan⁺¹(8.23/13.90) = 30.63° above horizontal ( Since initial vertical velocity is in upward direction)

   So Initial velocity = 16.15 m/s, 30.63° above horizontal