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
[tex]v_x = 28 m/s[/tex]
and the distance covered along the horizontal direction in a time t is
[tex]d_x = v_x t[/tex]
- The vertical motion of the skier is a uniformly accelerated motion, with initial velocity [tex]u_y = 0[/tex] and constant acceleration [tex]g=9.8 m/s^2[/tex] (where we take the downward direction as positive direction). Therefore, the vertical distance covered in a time t is
[tex]d_y = \frac{1}{2}gt^2[/tex]
The time t at which the skier lands is the time at which the skier reaches the incline, whose slope is
[tex]\theta = 36^{\circ}[/tex] below the horizontal
This happens when:
[tex]tan \theta = \frac{d_y}{d_x}[/tex]
Substituting and solving for t, we find:
[tex]tan \theta = \frac{\frac{1}{2}gt^2}{v_x t}= \frac{gt}{2v_x}\\t = \frac{2v_x}{g}tan \theta = \frac{2(28)}{9.8} tan 36^{\circ} =4.15 s[/tex]
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:
[tex]d_x = v_x t = (28)(4.15)=116.2 m[/tex]
[tex]d_y = \frac{1}{2}gt^2 = \frac{1}{2}(9.8)(4.15)^2=84.4 m[/tex]
The distance covered along the slope is just the magnitude of the resultant displacement, so we can use Pythagorean's theorem:
[tex]d=\sqrt{d_x^2+d_y^2}=\sqrt{(116.2)^+(84.4)^2}=143.6 m[/tex]
