Projectile motion skateboarder problem

[Chiborino]

Homework Statement

A skateboarder starts up a 1.0-m-high, 30-degree ramp at a speed of 7.0 m/s. The skateboard wheels roll without friction. How far from the end of the ramp does the skateboarder touch down?

angle = 30 degrees
v_i = 7.0 m/s
a = -9.8 m/s²
initial height = 1 m

Homework Equations

y = x*tanθ + ax²/(2*v_i^2*cos2θ) [graphs the path of a projectile]

v_f=v_i+at
v_f²=v_i²+2ad
d_y=v_iyt+.5at²
d_x=v_ixt

The Attempt at a Solution

First, I graphed the path of the projectile, and since it forms a parabola whose roots are the max distances, I got an answer of 5.65 m. The homework problem is online, so it told me that it was wrong.

Not knowing where I went wrong graphing it, I found the time it spent in the air by plugging into v_f²=v_i²+2ad, and then plugging v<sub>f into vf=vi+at for t0 to the max height. Then I just used the diy= .5at<sup>2 equation to find the time form the peak to the ground. The total time I got was .933 seconds. Multiplying this by the horizontal component of his velocity (vi*cosθ = 7.0*cos30 = 6.1) I got 5.69 seconds (probably a little different from my calculator since my calculator ignores significant figures even more than I do). Again, this was wrong.

So I gave up and clicked show answer: 3.78 meters.
I don't get how my teacher got this answer.</sup></sub>

 

[kuruman]

What did you use for the initial speed for the projectile part of the motion? I hope it was not 7 m/s because that's his velocity at the bottom of the ramp. At the top of the ramp, when he is about to take off, his velocity will be less than 7 m/s and you need to calculate what it is.

 

[Chiborino]

Ah, all the other problems in the set listed vi as the velocity the object becomes airborne with, so I assumed that was the case here as well. Factoring in that negative acceleration up the ramp got me 5.4 m/s leaving the ramp, which got me 3.78 meters.