Mechanics question. Person jumping off trampoline.

[Educated]

Homework Statement

Alex jumps off the edge of a trampoline at an angle of 40 degrees to the horizontal. The trampoline is 0.85 meters above the ground and Alex is in the air for 0.64 seconds.
(I did a rough sketch of it)

Calculate how far away from the edge of the trampoline Alex lands.
Neglect air resistance.

Homework Equations

The simple kinematic equations:
d = v_i t + \frac{1}{2}at^2

v_f = v_i + at

v_f \, ^2 = v_i \, ^2 + 2ad

d = \frac{(v_i + v_f)t}{2}

The Attempt at a Solution

I don't even know where to begin. The part that really confuses me is that the upwards part of the travel is shorter than the downwards part of the travel. I cannot solve the initial velocity, final velocity nor the distance.

Can someone just give me a hint in the right direction. I don't want the answer, leave me to solve the answer myself.

 

[Nynex]

Let the y traveled equal to -0.85...express x in terms of velocity...?

 

[Educated]

I'm sorry, what do you mean?
This question is really confusing me.

What is y and what is x? What will expressing things in terms of x and y achieve?

Mechanics really isn't my strong point.

 

[Nynex]

Have you guys studied projectile motion in class? The equations in your original post are based on a single dimension of displacement. In this problem Alex moves in two dimensions, those equations won't work!

Essentially, y would be the displacement in the y direction (up and down) and x would be the displacement in the x direction (left and right).

Alex displaces to the right from his original position. He also displaces down 0.85 m, since he lands on the ground below his starting position.

Using these two equations we should be able to solve for Alex's displacement in the x direction.
(sorry unfamiliar with latex)

x = (v*cos O)*t
y = (v*sin O)*t - g/2*t^2