Projectile motion: calculating time and distance

[janelle1905]

Homework Statement

A ski jumper travels down in a slope and leaves the ski track, moving horizontally with a speed of 25m/s. The landing incline below her falls off with a slope of 33 degrees.
a. How long is the ski jumper airborne?
b. Where does the ski jumper land on the incline?

Homework Equations

v = v₀ + at
x = x₀ + v_xt + 1/2 a_xt²

The Attempt at a Solution

a.
v_xo=25m/s
a_y=9.80m/s2

tan33^o=v_y/v_x=v_y/25
v_y0=24m/s

0=16.24 + (9.80)t
t=-1.7seconds***
***I'm not sure what I'm doing wrong that I'm getting a negative value for time

b.
y=0 + 0 + 1/2(9.80)(1.7)2
y=14
(I think this number should be a negative, since y₀=0 and the skier is going downhill?)

v_x=v_x0 + at
a=14.7

x=0 + 25(1.7) + 1/2(-14.7)(1.7)²
x=21

Thanks for your help :)

 

[kuruman]

The problem with your solution is that you set v_y/v_x equal to the tangent of the slope. v_y/v_x is the tangent of the angle at which the skier hits the slope at the end of the parabolic trajectory. Draw yourself the trajectory and include in your drawing the total horizontal distance the skier travels and the vertical drop. Do you see what is going on?

Also, your equations show an acceleration in the x-direction. There is no such thing.

 

[Jebus_Chris]

What you should do is draw yourself a picture. You have a flat surface which then slopes downward at $$33^o$$ and the skier is launching horizantally at the vertex with a speed of 25 m/s.

<< solution deleted by berkeman >>

 

[berkeman]

What you should do is draw yourself a picture. You have a flat surface which then slopes downward at $$33^o$$ and the skier is launching horizantally at the vertex with a speed of 25 m/s.

<< solution deleted by berkeman >>

Please do not do the OP's work for them. Provide tutorial hints, not full equations and solutions.