A ski jumper leaves the ski track in the horizontal direction with a speed of 25

Question

A ski jumper leaves the ski track in the horizontal direction with a speed of 25 m/s. The landing incline below him falls off with a slope of 35 degrees. Where does he land on the incline?

Explanation / Answer

from s = ut + (at^2)/2: horizontally, x = 25t => t = x/25 vertically y = -(gt^2)/2 as u = 0 so y = -(gx^2/625)/2 = -gx^2/1250 If the slope is angled at 35 degrees, the gradient is tan(35) = 0.7 approx formula of slope y = -0.7x Find point of intersection: -0.7x = -gx^2/1250 -0.7 = -gx/1250 x = 1250 * 0.7 / 9.81 = 89.2 approx y = 0.7 * 89.2 = -62.5 approx close enough? second part: if x is 89.2, from above: t = 89.2 / 25 = 3.57 s third part: from v = u + at v = 0 + gt v = 3.57 * 9.81 = 35 m/s

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