Question

a skier speeds along a flat patch of snow, and then flies horizontally off the edge at 16.0 m/s. He eventually lands on a straight, sloped section that is at an angle of 45.0∘ below the horizontal. How long is he in the air?

Answer 1

Answer:

1.63 s

Explanation:

The skier lands on the sloped section when the direction of its velocity is exactly identical to that of the slope, so at $45.0^{\circ}$ below the horizontal.

This occurs when the magnitude of the vertical velocity is equal to the horizontal velocity (in fact, $\tan \theta =\frac{|v_y|}{v_x}$, and since $\theta=45.0^{\circ}, tan \theta = 1$ and so $|v_y| = v_x$.

We already know the horizontal velocity of the skier:

$v_x = 16.0 m/s$

And this is constant during the entire motion.

The vertical velocity instead is given by

$v_y = u_y + gt$

where

$u_y = 0$ is the initial vertical velocity (zero since the skier flies off horizontally)

g = 9.8 m/s^2

t is the time

Here we have chosen the downward direction as positive direction.

Substituting $v_y = 16.0 m/s$, we find the time:

$t=\frac{v_y}{g}=\frac{16.0}{9.8}=1.63 s$