Ski Jumper Speed and Distance

## Question:

A ski jumper starts from rest at the top of a ski jump which has
a vertical drop to the take off point of 46 m. The take off point
is 4 m above the top of a landing slope which is 30 degrees below
horizontal. Assuming that we can neglect friction and air
resistance, a real stretch here, determine his speed at the take
off point and assuming he leaves the take off point horizontally,
determine the distance down the landing slope from the foot of
the ski jump under the take off point to the point of the
skier's landing. The skier has a mass of 70 kg.
## Answer:

The skier's energy is all potential at the top of the jump.
When he gets to the take off point, 46 meters worth of
gravitational potential energy will have been converted to
kinetic energy. Potential energy is m*g*h, in this case 70 kg * 9.8 m/s^{2} * 46 m = 31556
Joules. We know that
1/2*70*v^{2}=31556 so
v^{2}=902 m^{2}/s^{2}, so v=30 m/s.
With a horizontal launch at 30 m/s, the horizontal position of
the skier as a function of time is given by Dh=30*t. The vertical
position is given by
Dv=1/2*9.8*t^{2}. The vertical distance to landing
as a function of time is 4+30*t*sin(30) =
19*t. When the vertical distance to landing is equal to
the vertical distance fallen, the skier lands. When 4.9*t^{2}=19*t, the jump is over. This
occurs at time t=19/4.9 = 3.88 seconds.
In that time the skier has gone 30*3.88 meters downrange, 116.4
meters. The distance to the point of impact measured along the
landing slope will be 116.4/cos(30) = 134
meters.

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