A ski jumper leaves the ski track moving in the horizontal direction with a spee

Question

A ski jumper leaves the ski track moving in the horizontal direction with a speed of18.0 m/s, as shown in the figure. The landing incline below her falls off with a slope of 35.0°. Where does she land on the incline?

Express the coordinates of the jumper as a function of time: Substitute the values of xf and yf at the (3) dcos = vxit landing point (4) -d sin =--gt2 Solve Equation (3) for t and substitute the result into Equation (4) ( d cos Solve for d: g cos 2(18.0 m/s) sin 35.0o (9.80 m/s2) cos2 35.0° = 56.49 Xt = d cos = d cos 35.0° = 46.27 Evaluate the x and y coordinates of the point at which the skier lands: Yr =-d sin =-d sin 35.0° =-32.4 Finalize Let us compare these results to our expectations. We expected the horizontal distance to be on the order of 100 m, and our result is indeed on this order of magnitude. It might be useful to calculate the time interval that the jumper is in the air and compare it to our estimate of 4 s MASTER IT HINTS: GETTING STARTEDI I'M STUCK Suppose a ski-jumper lands at a point 157 m down the incline of the track described in the example (with a horizontal take-off). What was her speed at take-off? x m/s Your response differs from the correct answer by more than 100%

Explanation / Answer

h =vfy2/2g

vfy = [2gh]^1/2 =[2x9.8x157]^1/2

vfy = 55.47 m/s

vfx = vix = =18 m/s

v =[ (vfx )^2 +vfy )^2 ] = [(55.47)^2+(18)^2]

v =58.3 m/s

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