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.

We have equation of motion , [tex]s= ut+\frac{1}{2} at^2[/tex], 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

[tex]30 = 13.90*t+\frac{1}{2} *0*t^2\\ \\ t=2.16 seconds[/tex]

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 [tex]m/s^2[/tex] 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 = [tex]\sqrt{13.90^2+8.23^2} =16.15m/s[/tex]

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