Projectile Motion - Solving for initial velocity w/o time

[Sungura]

Homework Statement

A motorcyclist wants to jump 20 cars placed side-by-side, making the jump 30m long (coming back to the initial height). Assume that the ramp is placed at an angle of 30 degrees from the horizontal.

a. Calculate the initial velocity required to make the jump.

b. Assuming the motorcycle bounces elastically on the ground (rebounding at the same angle and speed), calculate the change in momentum of the motorcycle (mass = 500kg)

c. If the motorcycle is in contact with the ground for 0.25s, find the average force exerted by the ground on it.

Homework Equations

x=x0+v0t+(1/2)at^2
y=y0+v0t−1/2gt^2

The Attempt at a Solution

I'm not entirely sure if there are more equations I should be using, but I began with trying to use the x equation to solve for time. I assumed a was zero, since there is no acceleration in the x direction, and that x0 was zero, since we start at the origin. So, ending up with x=v0cosθ*t (Right here I'm not sure where the cos comes from, I kind of just stuck it there because I remember seeing such an equation before) I solved for t, ending up with t=30/v0cos(30).
I plugged this t back into the x equation, but I kind of get lost in the math at that point, since I seem to have to pull the v0 out of the right side of the equation somehow.

part (a.) seems to be my biggest problem, so if I get a little assistance with that I think it'll make (b.) and (c.) much easier to solve, since it's just impulse and momentum.

 

[Doc Al]

So, ending up with x=v0cosθ*t

You'll need this.

(Right here I'm not sure where the cos comes from, I kind of just stuck it there because I remember seeing such an equation before)

The initial velocity has horizontal (x) and vertical (y) components. Those components are handled differently. The cos comes from taking the x-component of the initial velocity.

You have an equation for the x-component of velocity. What about the y-component? Hint: How long will the motorcycle be in the air?

 

[Sungura]

You have an equation for the x-component of velocity. What about the y-component? Hint: How long will the motorcycle be in the air?

I just tried drawing out a rectangle with v0 as the hypotenuse and x and y as their respective sides, and found that y=v0sin(30).
I then tried to stick that into y=y0+v0t−1/2gt^2, but I got lost again once I ended up with v0sin(30)/t +.5gt^2=t

 

[Doc Al]

I just tried drawing out a rectangle with v0 as the hypotenuse and x and y as their respective sides, and found that y=v0sin(30).

Good. I think you mean $v_{0,y} = v_0 sin30$.

I then tried to stick that into y=y0+v0t−1/2gt^2, but I got lost again once I ended up with v0sin(30)/t +.5gt^2=t

You're using an equation for distance as a function of time. Try using an equation for velocity as a function of time?

 

[Sungura]

You're using an equation for distance as a function of time. Try using an equation for velocity as a function of time?

The only equations I can really think of are
v = v0 + at
v^2= v0^2 + 2a(x - x0)
But even plugging in the equations I simplified, I can't seem to get anywhere.

 

[Doc Al]

The only equations I can really think of are
v = v0 + at

That's the one you need. Use it to solve for the time that the motorcycle is in the air. (In terms of $v_0$.)