# Homework Help: Kinematics: How fast was he when he got off the ramp

1. Dec 6, 2009

### ramb

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

A bike is jumping between two ramps. The ramps are a height H , both with angles $$\theta$$ and seperated by a distance D. If he landed halfway down the landing ramp find the speed at which he left the launching ramp in terms of H, $$\theta$$ and D.

2. Relevant equations

1/2 * at^2 + vi*t + di = df
(i'm sure you all know that one)

3. The attempt at a solution

Okay so here's what I did, but apparently it was completely off, but i don't know where to begin it. *see attachment*

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2. Dec 6, 2009

### denverdoc

I like that eqn.

We know the takeoff angle. D is straightforward, how do we express the landing point in terms of the variables given. We are told it is 1/2 way down the ramp, so what is df and di. What are the initial x and y velocities?

How can we figure the time of flight?

3. Dec 6, 2009

### ramb

The landing point is:

$$\frac{H}{tan(\theta}$$

Let
$$L$$ = $$\frac{H}{tan(\theta}$$

With the equation
$$\frac{1}{2}at^{2}+v_{i}t+d_{i}$$

I separated all of it to x and y components and got:

$$v_{i}t=-\frac{at^{2}}{2}+d_{f}$$
then

$$v_{i}=-\frac{at^{2}}{2t}+\frac{d_{f}}{t}$$

so

$$v_{i}_{x}=-\frac{at}{2}+\frac{D+\frac{L}{2}}{t}$$

and

$$v_{i}_{y}=-\frac{gt}{2}-\frac{H}{2t}$$

Once I have both of them, do i just have to square both of them, put them under the square route then, finish?

$$v_{i}=\sqrt{(-\frac{at}{2}+\frac{D+\frac{L}{2}}{t})^{2}+(-\frac{gt}{2}-\frac{H}{2t})^{2}}$$

then substituting for L
$$v_{i}=\sqrt{(-\frac{at}{2}+\frac{D+\frac{(\frac{H}{tan(\theta)})}{2}}{t})^{2}+(-\frac{gt}{2}-\frac{H}{2t})^{2}}$$

Last edited: Dec 6, 2009
4. Dec 6, 2009

### denverdoc

I'll need a sec to look at this, just got back. Stay tuned.

5. Dec 6, 2009

### denverdoc

here is the approach I have taken, so far unfruitful cuz of the 1/2 ramp displacement which bungs up what is usually very clean algebra: let a = angle of the ramp.

Delta x= V Cos(a)*t=.5H/tan(a) + d

Delta y = 1/2H=1/2 gt^2+ V sin(a)*t (a quadratic in t but only the longer time will work)

Normally what I do here would be substitute for t which leads to:

let d+0.5H/tan(a)= C; then 1/2H=1/2 g (C/(V*cos(a)))^2 + (V sin(a)*C/V cos(a))
" = " + C (tan(a))
There may be a trig ID here which can be uses to simplify things considerably.

6. Dec 6, 2009

### ramb

Thanks,

so why did you substitute the t, and where did you get the C from?

7. Dec 6, 2009

### ramb

oh I see,

so with the time it took in the x direction, you used that time in terms of v,d,h,theta to put it in the y equation, since that's the one where the time in air matters. So after all that, the only thing is solve for V?