Question

A ski jumper travels down a slope and leaves

the ski track moving in the horizontal direc-
tion with a speed of 28 m/s as in the figure.
The landing incline below her falls off with a
slope of 0 = 36º.
The acceleration of gravity is 9.8 m/s.

Answer 1

Question seems to be missing. Found it on google:

a) How long is the ski jumper airborne?

b) Where does the ski jumper land on the incline?

a) 4.15 s

We start by noticing that:

- The horizontal motion of the skier is a uniform motion, with constant velocity

`v_x = 28 m/s`

and the distance covered along the horizontal direction in a time t is

`d_x = v_x t`

- The vertical motion of the skier is a uniformly accelerated motion, with initial velocity
`u_y = 0` and constant acceleration
`g=9.8 m/s^2` (where we take the downward direction as positive direction). Therefore, the vertical distance covered in a time t is

`d_y = (1)/(2)gt^2`

The time t at which the skier lands is the time at which the skier reaches the incline, whose slope is

`\theta = 36^(\circ)` below the horizontal

This happens when:

`tan \theta = (d_y)/(d_x)`

Substituting and solving for t, we find:

`tan \theta = ((1)/(2)gt^2)/(v_x t)= (gt)/(2v_x)\\t = (2v_x)/(g)tan \theta = (2(28))/(9.8) tan 36^(\circ) =4.15 s`

b) 143.6 m

Here we want to find the distance covered along the slope of the incline, so we need to find the horizontal and vertical components of the displacement first:

`d_x = v_x t = (28)(4.15)=116.2 m`

`d_y = (1)/(2)gt^2 = (1)/(2)(9.8)(4.15)^2=84.4 m`

The distance covered along the slope is just the magnitude of the resultant displacement, so we can use Pythagorean's theorem:

`d=√(d_x^2+d_y^2)=√((116.2)^+(84.4)^2)=143.6 m`