Question

A ski jumper travels down a slope and leaves

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

$d_y = \frac{1}{2}gt^2$

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:

$tan \theta = \frac{d_y}{d_x}$

Substituting and solving for t, we find:

$tan \theta = \frac{\frac{1}{2}gt^2}{v_x t}= \frac{gt}{2v_x}\\t = \frac{2v_x}{g}tan \theta = \frac{2(28)}{9.8} tan 36^{\circ} =4.15 s$

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:

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

$d_y = \frac{1}{2}gt^2 = \frac{1}{2}(9.8)(4.15)^2=84.4 m$

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

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(116.2)^+(84.4)^2}=143.6 m$