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?

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

3is a crest4 is a trough 2 is the amplitude 1 is the wavelength