Projectile motion skateboarder help

This problem is not fun:
Code:
A skateboarder starts up a 1.0-m-high,
30deg ramp at a speed of 7.0 m/s.
The skateboard wheels roll without friction.

I guees the first thing I'll ask is the 7 m/s at 30 deg the launch velocity and angle?

Doc Al
Mentor
No, 7 m/s is the speed at the bottom of the ramp. You'll need to figure out the speed when it takes off at the top.

ok, I don't know how to resolve this. I understand that the skateboarder is going to be slower at the top and gravity is the only force stopping motion.

Doc Al
Mentor
If you want to find the speed at the top, use conservation of energy. Or find the acceleration up the ramp.

I have not learned the conservation of energy yet. I just used the formula sqrt(2*g*sin(theta)*delta(x)
which is
sqrt(2*9.8*sin(30)*1.73=4.11 m/s^2

I still get the answer wrong.

Homework Helper
NIZBIT said:
I have not learned the conservation of energy yet. I just used the formula sqrt(2*g*sin(theta)*delta(x)
which is
sqrt(2*9.8*sin(30)*1.73=4.11 m/s^2

I still get the answer wrong.

Use v = Vo + at, where Vo is your initial speed, and a the acceleration. Now, what does this acceleration equal while the skater is climbing up the ramp?

I'm guessing -9.8? But I dont know the time so how can I use that?

Doc Al
Mentor
NIZBIT said:
I have not learned the conservation of energy yet. I just used the formula sqrt(2*g*sin(theta)*delta(x)
which is
sqrt(2*9.8*sin(30)*1.73=4.11 m/s^2

I still get the answer wrong.
What's the 1.73 signify?

What you are using here--and which is perfectly OK if done right--is the formula:
$$v^2 = v_0^2 + 2 a \Delta x$$

To use it properly, answer these questions:
What's the acceleration along the ramp?
What's the displacement along the ramp?
What's the initial velocity?

Well honestly, I 've done this problem so much I really don't know anymore. I dont know if accelreation is 9.8 or -gsin(theta). The displacement I'm assuming is the hypotenuse which is 2.

You're right, take g sin theta for the acc (remember it's negative) and the displacement along the ramp is 2.

Ok so I got 6.26 m/s. Now the problem becomes projectile motion I hope. Now I just need to find the range.

That's not what I get.....can you show me your calculation.

As far as I can see, both your distance up the slope and the acceleration with have sin30 in them, so the sin30s will cancel.

I did sqrt((7^2)+(2*-9.8sin(30)*2)).

Homework Helper
NIZBIT said:
I did sqrt((7^2)+(2*-9.8sin(30)*2)).

Seems correct, as far as I'm concerned.. I did another 'method' too, to check.

NIZBIT said:
I did sqrt((7^2)+(2*-9.8sin(30)*2)).

That gives me 49 - 19.6 which gives me 29.4. The sq rt of that is not 6.26

Homework Helper
rsk said:
That gives me 49 - 19.6 which gives me 29.4. The sq rt of that is not 6.26

$$\sqrt{7^2 - 2\cdot9.81\cdot\sin(30)\cdot2} \approx 5.42$$.

(...Which perfectly fits the expression Doc Al gave: $$v^2 = v_0^2 + 2 a \Delta x$$.)

And which is why I said I didn't agree with 6.26 m/s