Question

A ski jumper leaves the ski-track moving the horizontal direction with a speed of 25m/s. the landing incline below her falls off with a slope of 35®. Where does she land on the incline?

Answer 1

Given,

The speed of a ski jumper, v = 25 m/s

The angle of slope = 35 ^o

By using s = u t + (1/2) at²

The horizontal direction, x = 25t => t = x/25

The vertical direction y = -(g t²)/2 as u = 0

so y = -(g x²/625) / 2 = -g x² / 1250

If the slope is angled at 35 ^o, the gradient is tan(35) = 0.7

The formula of slope y = - 0.7x

The point of intersection is,

-0.7x = -g x² / 1250

-0.7 = -g x/1250

x = 1250 * 0.7 / 9.81 = 89.2

y = 0.7 * 89.2 = - 62.5

If x is 89.2, from above:

t = 89.2 / 25 = 3.57 s

From v = u + at

v = 0 + gt

v = 3.57 * 9.81

= 35 m/s