A skier skis from rest from a vertical height h1 = 17.3 m over two successively

Question

A skier skis from rest from a vertical height h1 = 17.3 m over two successively lower hills of vertical heights h2 = 15 m and h3 = 7.9 m. (Fig. 7-30). The summit of the third hill fits a circle of radius h3 centered at height 0 m. Friction with the snow and air resistance are negligible. Figure 7-30 (a) Find her speeds at x1, x2 and X3. V1=0 m/s V2 =6.714 m/s V3= 13.57 m/s (b) Does the skier leave the surface at x3? Yes No If not, what should h1 be so that she just leaves the surface at x3? If yes, find the maximum value of h1 at which the skier stays on. (Ignore the middle hill in this calculation.) m

Explanation / Answer

Since the path of the skier is of a circle, for the skier to leave the surface,

(v^2)/r > g

the centripetal acceleration is g which must be less than the centrifugal acceleration the skier feels from his frame
of reference.

taking g=9.8 and r=7.9 we get the value of v=8.798 m/s
since the skier already has 13.57 m/s in our case he leaves the surface at x3.

The height of h1 must thus be reduced so that the skier stays on the surface.
assume h1=X
then, V3=sqroot(2xgx(h1-h3))
which must be less than sqroot(h3xg)

equation is: sqroot (2xgx(h1-h3)) < sqroot(h3xg)

solving this 2xgx(h1-h3)<h3xg
   2x(h1-h3)<h3
   h1<7.9+3.95
   h1<11.85m

But this might not help us cross the middle hill if it is there because we need atleast 15m height to cross it.
So the middle hill must not be there if we want the skier to stay on ground at h3.

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