How Far Does a Skier Jump at 15° Launch and 50° Slope?

[Jbright1406]

Homework Statement

A skier leaves the ramp of a ski jump with a velocity of v = 8.0 m/s, θ = 15.0° above the horizontal, as in the figure. The slope is inclined at 50.0°, and air resistance is negligible. (Assume up and right are positive, and down and left are negative.)

(a) Find the distance from the ramp to where the jumper lands.

Homework Equations

The Attempt at a Solution

what I am coming up with is
dcos50=(8cos15)t deltaX=Vxi(t)
-dsin50=(8sin15)t+.5(-9.8)t^2 deltaY=Vyt+.5(g)t^2

im stuck at how to solve for d and t. and i have to have this done by 11:30 :-) any helps on cracking this is more than appreciated. i think I am supposed to solve for one of them and then plug it into the other equation but I am not sure how to do that

 

[Nabeshin]

Try solving for time in the y direction and substituting into the x equation to obtain a distance. This will involve solving a quadratic.

 

[Jbright1406]

ok, my times up anyways but i still want to figure it out. how do i solve for t in the y direction. cause in m equation i still have d right? i know the quadratic formule is x equals negative b plus or minus the sqr root of b^2-4ac all divided by 2a. i get lost there

 

[rl.bhat]

Wright the projectile equation as

y = [tan(theta)]x - g*x^2/[(Vocos(theta)]^2
Substitute x = d*cos(theta) and d*sin(theta) and solve for d.

 

[Jbright1406]

huh? i have alpha and beta. alpha is 15 beta is 50