Find the takeoff speed of the skier (Master It) A ski jumper leaves the ski trac

Question

Find the takeoff speed of the skier (Master It)

A ski jumper leaves the ski track moving in the horizontal direction with a speed of 24.0 m/s, as shown in the figure. The landing incline below her falls off with a slope of 35.0 degree. Where does she lard or the incline? Conceptualize We can conceptualize this problem based or memories of observing winter Olympic ski competitions. We estimate the skier to be airborne for perhaps 4 s and to travel a distance of about 100 m horizontally. We should expect the value of d, the distance traveled along the incline, to be of the same order of magnitude. We categorize the problem as one of a particle in projectile motion. It is convenient to select the beginning of the jump as the origin. The initial velocity components are v_xi = 24.0 m/s ard v_yi = 0. From the right triangle in the figure, we see that the jumper's x and y coordinates at the larding point are given by xf = d cos 35.0 degree and yf = -d sin 35.0 degree. 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 or 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.

Explanation / Answer

equation for d is

d = 2*Vxi^2*sin(35)/(g*cos^2(35))

we can re write as

d1/d2 = vxi1^2/vxi2^2

100.4/110= 24^2/vxi2^2

vxi2 = 25.12 m/s

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