A skilled skier knows to jump upward before reaching a downward slope. Consider

Question

A skilled skier knows to jump upward before reaching a downward slope. Consider a jump in which the launch speed is v_0 = 10.5 m/s, the launch angle is theta_0 = 8.2 degree, the initial course is approximately flat, and the steeper track has a slope of 12.5 degree. Figure (a) shows a prejump that allows the skier to land on the top portion of the steeper track. Figure (b) shows a jump at the edge of the steeper track. In Figure (a), the skier lands at approximately the launch level, (a) In the landing, what is the angle phi between the skier's path and the slope? In Figure (b), (b) how far below the launch level does the skier land and (c) what is phi? (The greater fall and greater phi can result in loss of control in the landing.)

Explanation / Answer

(a)In the landing, the angle between the skier's path and the slope = -8.2 degrees

We assume, in a coordinate system, that the skier jumps from the origin, and that the slope turns downward at the origin.

The equation of the downward slope is y = -tan12.5º x (Eq 1).

The equation of horizontal displacement is x = 10.5 cos 8.2º t (Eq 2).

Therefore the y-value of the landing spot, as a function of time, is y = -10.5 cos 8.2º tan 12.5º t (Eq 3).

The equation of vertical displacement is

y = 1/2 gt^2 + 10.5 sin 8.2º t = -4.9 t^2 + 10.5 sin 8.2º t (Eq 4).

Solve Eq 3 & 4 for t and y:

-10.5 cos 8.2º tan 12.5º t = -4.9 t^2 + 10.5 sin 8.2º t ==> 4.9 t = 10.5 (sin 8.2º + cos 8.2º tan 12.5º) ==>

t = (10.5/4.9) (sin 8.2º + cos 8.2º tan 12.5º) = 0.7758s

Plug this t-value into Eq. 3 to get your answer (b).

y = -10.5 cos 8.2º tan 12.5º (0.7758) = -1.788m

To get answer (c), we need the horizontal and vertical velocity components at time t. The horizontal component is constant, vx = 10.5 cos 8.2º. The vertical component is the derivative of Eq 4, evaluated at time t:

vy = -9.8 t + 10.5 sin 8.2º

Now, tan = vy/vx = -0.98 t / cos 8.2º + 10.5 tan 8.2º => angle = -16.89 degrees

gives you the angle between the skier's path and the horizontal. To get the angle between the path and the slope, = || - 11.3º

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