Question

Eddie the eagle, british olympic ski jumper, is attempting his most mediocre jump yet. after leaving the end of the ski ramp, he lands downhill at a point that is displaced 72.1 m horizontally from the edge of the ramp. his velocity just before landing is 33.0 m/s and points in a direction 30.0$^\circ$ below the horizontal. neglect any effects due to air resistance or lift. what was the magnitude of eddie's initial velocity as he left the ramp?

Answer 1

When he lands his horizontal velocity is 
28 cos40 = 21.45 m/s 

the time in flight comes from 
x = Vht 
58.8 = 21.45t 
t = 2.74 seconds 

his vertical velocity at landing is 
28 sin40 = -18 m/s 

his vertical velocity equation is 
v = V0 - gt 
-18 = V0 - 9.81(2.74) 
V0 = -18 + 9.81(2.74) 
V0 = 8.88 

his velocity magnitude was 
v = (8.88^2 + 21.45^2)^½ 
v = 23.2 m/s ANSWER 

his initial direction was 
tanθ = 8.88/21.45 
θ = 22.5 degrees above the horizontal ANSWER 

to find the time to the flight apex from launch 
v = gt 
8.88 = 9.81t 
t = 0.905 s 

in 0.905 s Eddie has risen how far above the edge 
y = ½(9.81)(0.905^2) 
y = 4 m 

the remainder of the flight is all drop and takes 2.74 - 0.905 = 1.85 seconds 

in 1.85 seconds he drops 
y = ½(9.81)(1.85^2) 
y = 16.7 m 

so the height from the edge to the landing point is 
16.7 - 4 = 12.7 m ANSWER