# Skating ramp, only height given. Kinematic equations.

1. Oct 26, 2015

### LogarithmLuke

1. The problem statement, all variables and given/known data
A skater jumps off a quarterpipe ramp and reaches 5m over the ramp. We can assume that he jumps in a vertical line. Find the speed the skater had before jumping off the ramp.

2. Relevant equations
All the 1-D kinematic equations. It is not intended that you should use any more advanced equations than this.

3. The attempt at a solution
It may seem like i haven't put in any effort, but i really have no clue as to how to solve this one. We have only been given the height, no speed, no acceleration and no start speed. I have never solved a problem like this where i've been given so little information.

2. Oct 26, 2015

### BvU

And yet it is solvable ! A matter of picking the right equations from Doc's long list. Your job ! Propose a few candidates and we'll help you find the right one(s).

3. Oct 26, 2015

### SteamKing

Staff Emeritus
You've been given all the information necessary to solve this problem.

Lookit:
1. A skater jumps off a quarterpipe ramp and reaches 5m over the ramp. This tells you how high the skater was able to go after leaving the ramp
2. We can assume that he jumps in a vertical line. This tells you what kind of motion to use.

So, to recap: something travels in a vertical line for 5 meters after jumping from something.

Could you envision a solution if I told you that instead of someone jumping, you threw a ball straight up 5 meters. Find out how fast the ball was thrown to reach this distance?

4. Oct 26, 2015

### LogarithmLuke

I would pick one that includes length/position as this is the only variable we know. I couldn't find it on the list, but at school we used the following equation alot, where x is the position/length. x=1/2(v0+v)t. I guess you could also consider the displacement equation Doc AL listed: x=x0+v0t+(1/2)at^2.

So if it was a ball we know that it has the acceleration 9.81m/s^2 when falling down vertically, but i still wouldnt be able to solve it just knowing the height.

5. Oct 26, 2015

### BvU

Good start. x=1/2(v0+v)t is nice for an average velocity 1/2(v0+v). It may even work here. x=x0+v0t+(1/2)at^2 looks a bit more promising ! You know a, you know x - x0 and that leaves two unknown variables. So we need one more equation... must be in the vicinity of this one in Doc's list !

"length/position as this is the only variable we know" is not completely true. What about the vertical velocity at the highest point ?

6. Oct 26, 2015

### LogarithmLuke

Hm so we know that v(end speed)=0 when s(height)=5. Perhaps we could use this one: v^2=v0^2+2as. We know v(when s=5). We know a, and we have a value to use for s. Seems like that one should work.

7. Oct 26, 2015

### BvU

I would think so ...
A safer way to bring in this equation is as "conservation of mechanical energy": $\ \ {1\over 2} m v_0^2 + mg h_0 = {1\over 2} m v_1^2 + mg h_1 \ \$ which then becomes your equation.

Another candidate is $v(t) = v(0) + at$

For the answer, it of course boils down to the same.