# Projectile Motion skateboarder

1. Sep 28, 2014

### Physicsnoob90

1. The problem statement, all variables and given/known data

A skateboarder shoots off a ramp with a velocity of 5.30 m/s, directed at an angle of 52.0° above the horizontal. The end of the ramp is 1.40 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp.

(a) How high above the ground is the highest point that the skateboarder reaches?
(b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

2. Relevant equations

3. The attempt at a solution

My setup

a = -9.9m/s^2
Vxo= Vocosø
Vyo= Vosinø

My steps:
1) used Vy^2 = Vyo^2+2ayY and had Vy = 0, my answer was 0.08899m

I don't know if i did this correct/right. Can anyone help me?

2. Sep 28, 2014

### MrPunk44

That answer seems to be off by an order of magnitude. I would recheck your calculation. Also, the height that you calculated there is NOT the height the question is asking for. Can you tell the difference between the height you calculated and the height the question wants?

3. Sep 28, 2014

### Physicsnoob90

Would the height of the (a) be 1.40 m + x since the height of the ramp is 1.40 m above the ground? Also, i'm a little confuse on how to find the magnitude of this problem.

4. Sep 28, 2014

### MrPunk44

Yes, that is correct.

I'm unsure what you mean by magnitude. I was referring to your answer of 0.08899m. That answer is off by one factor of 10. I assume it was just an error when you entered in the value?

5. Sep 28, 2014

### Physicsnoob90

Also, do Vy=0? since the y direction is pointing upward

6. Sep 28, 2014

### MrPunk44

Vy means the velocity in the y direction. In this problem it equals 0 at one point in time and that is when it reaches its maximum height.

7. Sep 28, 2014

Ok thanks.