Aski 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.

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

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

- The vertical motion of the skier is a uniformly accelerated motion, with initial velocity and constant acceleration (where we take the downward direction as positive direction). Therefore, the vertical distance covered in a time t is

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

below the horizontal

This happens when:

Substituting and solving for t, we find:

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:

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

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer