Mechanical Energy help?

Question

http://i21.photobucket.com/albums/b272/Cajunboiler/06_45.gif

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. As the drawing shows, one person hits the water 5.00 m from the end of the slide in a time of 0.500 s after leaving the slide. Ignoring friction and air resistance, find the height H in the drawing in meters.

Now, I know this has to do with the conservation of mechanical energy. For this problem, I think it should be mg(ho) = .5m(vf)^2

Now... I'm thinking I need to use kinematics to find vf but I'm confused on which equation to use.

If this is not how to work it, the please explain it to me where I went wrong.

I keep getting 5.1 m for the height, but my online homework website is telling me that it is wrong. I'm getting 10 for vf, so that's obviously wrong.

Does the height where the person leave the slide matters? The H is the height from the top of the slide to the water in the picture.

OOOOOOOOOOOO, I get it now.

What I was solving for was only one part of the problem, 5.1. The other part was adding 1.226. Thanks for the help everyone!

Answer 1

You are correct. You have a good start.

The vf, velocity at the end of the slide will only have an x component.

How we will tackle this is figure out the vf_x, the x component of the vf.

We do not know h_f, the height of the bottom of the slide to the water. We need this in order to find vf, from there it's a piece of cake.

In this case, our main equation is the distance & time kinematic equation:

d = v_0 * t + (1/2) a * t^2

v_0 in our case, is the vertical velocity the person leaves the slide at.

v_0 = v_fy = 0 m/s since it is horizontal

a = 9.81 m/s

t = .5

d = h

so

h_f = 0 + 0.5 * 9.81 * (.5^2)


h_f = 1.226 meters

=========
The next step is to find v_fx. This is simple.

Recall that the horizontal component of velocity will not change since we are disregarding drag. Gravity only works on the vertical component of velocity.

Thus

v_fx = 5.00 m/ .500 sec
v_fx = 10 m/s
v_f = 10 m/s

========
Now we need to use our mechanical energy equation.
mg(ho) = .5m(vf)^2
ho = H - h_f since our v_f is completely derived from this distance.

so

mg(H - h_f) = .5m(v_f)^2
9.81(H - 1.226) = .5 * 100

Solve for H.

H = 6.3228 meters

** update in response to your update
It depends on how you solve it. There are 2 main ways, one using the velocity at impact, and the other using the slide velocity.

I did it using the slide exit velocity. Mensajero has already done it correctly using the final velocity at impact. As you can see we arrived at the same answer (accounting for rounding and things).

If you use the velocity at impact, you will find H. If you use the slide exit velocity, you will need to subtract the bottom of the slide height.

Answer 2

ok, here's how i would do it:

let's say that H = a + b

where a = height of the slite and b= the height from the tip of the slide to the water

you know that m*g*a=0.5*m*(vf)2

since the mass is the same you get rid of it

g*a = 0.5*vf2

you don't know vf, so you have to get it

you suppose that he leaves the slide horizontaly so you get that he touches the water after 5m and it takes him .5s to reach it

you use this equation to get b

b= vot+0.5gtsquared

you know that vo in the y axis is going to be 0 so

b = 0.5g*tsquared

b = 0.5 * 9.81 m/s2 * (0.5)squared

b = 1.226m

since there is no air resistance and he leaves the slide horizontaly the velocity will be

Vo = Vf = 5m/0.5s

Vf = 10m/s

now you just use the energy equation

g*a = 0.5*vf squared

where a = 0.5 Vfsquared/g

a = 0.5 * (10 m/s)squared / 9.81 m/s

you get that

a = 5.1m

so H = a+b

H = 5.1m +1.226m = 6.33 m

hope this helps.

Answer 3

to find v(f) calculate horizontal and vertical component of velocity when it hits the water.then add them vectorally.