# Energy methods with no numbers

1. Oct 14, 2014

### Ering

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

Ally starts at rest with a height H above the ground and slides down a frictionless slide. The bottom of the slide is a height h above the ground. Ally then leaves the slide horizontally, striking the ground a distance d from the end of the slide (where she left the slide).
Use energy methods to help determine the initial height H of Ally in terms of h and d.

2. Relevant equations

Ei = Ef (initial energy equals final energy)

Ki + Ui = Kf + Uf (initial kinetic energy plus initial gravitational potential energy equals final ...)
--> 1/2 m(vi)^2 + mg(yi) = 1/2 m(vf)^2 + mg(yf)
m = mass
g = gravity
y = height
v = initial/final velocity

3. The attempt at a solution

I started by writing out:
Ki + Ui = Kf + Uf
Then, i figured since Ally starts at rest, the initial velocity is zero which would cancel out Ki. And since Ally lands on the ground, the final height (yf) would be zero so i cancelled the grav. potential final energy too.
So,
mg(yi) = 1/2 m(vf)^2

Since I want the initial height (yi), i rearranged the equation to solve for yi, and cancelled m's

yi = vf^2 / 2g

Then i figured I could substitute some equation in for vf^2 that incorporated the h and d... however i'm not sure which equation(s) i could use?
Probably something to do with projectile motion, but i'm really not sure where to go from here.

Thanks for any help

-stuck and stressed

2. Oct 14, 2014

### Simon Bridge

You need some sort of energy argument for ballistics.
At the end of the slide - Ally has lost some potential energy, which has gone into kinetic energy.
Lets call this K1:

$K_1=mg(H-h)=\frac{1}{2}mv^2$ where v is the speed that Alley leaves the ramp.

Consider - if t is the time from the end of the ramp to hitting the ground, then $d=vt$ right ... but we need an energy argument.
That means we need to make that equation look like an energy equation ... since it has a v in it, we can try for kinetic energy: square the equation, then multiply both sides by m/2:

$\frac{1}{2}md^2 = \frac{1}{2}mv^2t^2 = K_1t^2$

You also know an equation for the time it takes for Alley to fall distance h.
You can work out the energy form of that equation - and we'd hope that the non-energy terms will cancel out.
Then you will have learned an energy relation for ballistics.

Its possible you already have one or you can get it another way - I'm just trying to build on your current understanding.

3. Oct 14, 2014

### Ering

For the equation for the time it takes to fall distance h, would you use
Δy = Voy t + 1/2 a t2 ? where Δy = h

If so, would you set that equal to the equation you gave above? Or how do I make it so terms can cancel out?

4. Oct 14, 2014

### Ering

For the equation for the time it takes to fall distance h, would you use
Δy = Voy t + 1/2 a t2 ? where Δy = h

If so, would you set that equal to the equation you gave above? Or how do I make it so terms can cancel out?

5. Oct 15, 2014

### nasu

Vo is here the velocity at the bottom of the slide. It says that she leaves the slide horizontally. So what is Voy?
Then combine this with the equation for the horizontal motion (d=vox t) to eliminate the time and find Vo^2 as a function of h,d and g.
Once you have Vo^2 plug in the equation expressing conservation of energy between height H and height h.