# Projectile motion skateboarder help

1. Oct 26, 2006

### NIZBIT

This problem is not fun:
Code (Text):

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?

2. Oct 26, 2006

### Staff: 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.

3. Oct 26, 2006

### NIZBIT

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.

4. Oct 26, 2006

### Staff: Mentor

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

5. Oct 26, 2006

### NIZBIT

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.

6. Oct 26, 2006

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?

7. Oct 26, 2006

### NIZBIT

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

8. Oct 26, 2006

### Staff: Mentor

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?

9. Oct 26, 2006

### NIZBIT

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.

10. Oct 26, 2006

### rsk

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

11. Oct 26, 2006

### NIZBIT

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

12. Oct 26, 2006

### rsk

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.

13. Oct 26, 2006

### NIZBIT

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

14. Oct 26, 2006

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

15. Oct 26, 2006

### rsk

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

16. Oct 26, 2006

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

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

17. Oct 26, 2006

### rsk

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

18. Oct 26, 2006

Ooops, sorry, I seem to have misread you. :shy: