# Kinematics in two dimensions

1. Nov 6, 2004

### FancyNut

This problem looks so simple but every answer I get is wrong...

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?

EDIT : I realized several mistakes.. this is how I did it the last time:

I used the y-component of motion (free fall) to calculate time..

$$0 = 1 + v_0 sin (30) t- \frac {g}{2} t^2$$

then I plug into this:

$$x_f = v_0 cos (30) t$$

I have no idea what I'm doing wrong here... Ancy help would be much appreciated.

Last edited: Nov 6, 2004
2. Nov 6, 2004

### boaz

i think you chose the wrong y and x axises. i suggest you to choose the x axis directed along the ramp surface line. then you will have to separate the mg force to y and x components. it will help me a lot, if you also type the answer which is written in your workbook.

3. Nov 6, 2004

### FancyNut

I'll try that new coordinate system...

I don't have the final answer. I'm using www.masteringphysics.com and the only way to get it is to request it which will end the problem for me-- 10 points off. :(

4. Nov 6, 2004

### FancyNut

Ok I guess I should've calculated the final velocity the skater takes off..

acceleration is - 9.8 sin 30 and after doing everything again it's still.... wrong.

5. Nov 6, 2004

### boaz

ok, then we go to plan b. can you give me more details, such as the weight of the skater, how long is the ramp and etc?

6. Nov 6, 2004

### FancyNut

There is no weight given.

I just tried it again... this time I calculated how long the ramp is which is 2 (since hight is 1 and sin 30 = 1/2). Now using this distance I calculated the final velocity the dude leaves the ramp with... then using that velocity I took the y-component of it and calcuated the time it takes to hit ground... used that time to calculate final position... and it was wrong.

7. Nov 6, 2004

### FancyNut

I think I have it right this time except one thing... this second degree equation. I used the quadratic formual... and after applying it and using the new t value I got the answer is STILL wrong. Why does this have to happen.

$$0 = 1 + 14.7 t - \frac {g}{2} t^2$$

$$4.9 t^2 - 14.7 t - 1 = 0$$

Last edited: Nov 6, 2004
8. Nov 6, 2004

anybody... ?

9. Nov 7, 2004

### Skomatth

Where did u get 14.7 from?

To get the speed at the top of the ramp.

10. Nov 7, 2004

### Duarh

I recommend you use conservation of energy.

We have MVinitial^2/2=MVfinal^2/2+Mgh
V^2=Vfinal^2+2gh

where Vfinal - speed at top of ramp, g - acceleration of gravity - h height of ramp. once you have vfinal, you can calculate its y component by trig and see how long it'll take for the skater to go up and down to the earth again (under acceleration -g), and then use that time to calculate the horizontal distance from the horizontal component of Vfinal

11. Nov 7, 2004

### FancyNut

whew. I finally got the right answer. Thanks guys!

Is it me or is masteringphysics.com the scariest educational tool in existence? Typing in my number and pressing the 'submit' tab while dreading the 'try again' message is a horrible, horrible experience. :(