In an olympic trial, a ski jumper shoots off the end of a ramp with an initial velocity of 21.30 m/s, directed

at an angle of 35° above the horizontal. The end of the ramp is height h = 11.0 m above bottom of the hill as shown in the diagram above.

When the skier reaches the bottom of the hill, how far is this point horizontally from the end of the ski jump?

(c) What is the velocity (speed and direction) that the skier has when he lands at the bottom of the hill?
speed ______ direction __________ ° below the horizontal.

Horizontal component of launch velocity = (cos 35) x 21.3, = 17.45m/sec.
Vertical component = (sin 35) x 21.3, = 12.22m/sec.
Time to maximum height = (v/g), = 12.22/9.8, = 1.247 secs.
Maximum height from ramp = (v^2/2g), = (12.22^2/19.6), = 7.5939 metres.
Total height to drop = (7.5939 + 11m) = 18.5939 metres.
Time to drop from max. height = sqrt. (2h/g), = 1.948 secs.
Total time in air = (1.948 + 1.247) = 3.195 secs.
Distance horizontally = (3.195 x 17.45) = 55.75metres.

Vertical velocity at landing = sqrt. (2gh), = sqrt. (19.6 x 18.5939) = 19.09m/sec.
Horizontal component = 17.45m/sec.
Velocity = sqrt. (19.09^2 + 17.45^2), = 25.864m/sec., and angle = arctan (19.09/17.45), = 47.56 degrees from horizontal.