Question

After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 31 m horizontally from the end of the ramp. His velocity, just before landing, is 21.5 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

v_f =21.5m/s

v_fx=v_fcosθ = 21.5cos43 = 15.72m/s

v_fy=v_fsinθ = 21.5sin43 = -14.66m/s

Since there is no acceleration along horizontal direction a_x=0m/s²

v_fx= v_ix+a_x*t = v_ix+0*t

v_fx= v_ix= 15.72m/s

time required to traveled distance X=31m is = t

t=X/ v_fx= 31/15.72 = 1.97s

Acceleration along vertical direction = a_y=g=-9.8m/s²

v_fy= v_iy+a_y*t = v_iy - g*t

v_iy = v_fy+g*t

v_iy = -14.66 + 9.8*1.97

v_iy= 4.65m/s

v_i= sqrt[v_ix²+ v_iy²) = sqrt(4.65^2+15.72^2)

v_i= 16.39m/s

Φ = tan^-1(v_iy / v_ix) = tan^-1(4.65/15.72)

Φ = 16.5^o