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?

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

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

We already know the horizontal velocity of the skier:

[tex]v_x = 16.0 m/s[/tex]

And this is constant during the entire motion.

The vertical velocity instead is given by

[tex]v_y = u_y + gt[/tex]

where

[tex]u_y = 0[/tex] 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 [tex]v_y = 16.0 m/s[/tex], we find the time:

[tex]t=\frac{v_y}{g}=\frac{16.0}{9.8}=1.63 s[/tex]