# Velocity down an incline

## Homework Statement

A 75KG object, goes down a slide that is 10m tall. What is the speed of the object at the halfway mark and the speed at the bottom of the slide. The object starts at rest and there is no friction.

## Homework Equations

v2f - v2i = 2as

3. The Attempt at a Solution [/B]
attached pic of where I am at on the problem Related Introductory Physics Homework Help News on Phys.org
ehild
Homework Helper
You have the equation: v2f - v2i = 2as. Note that s is the displacement along the slope now. How long is it in terms of the angle if the height of the slope is 10 m?

You have the equation: v2f - v2i = 2as. Note that s is the displacement along the slope now. How long is it in terms of the angle if the height of the slope is 10 m?
sinθ=(10/h)
h equaling the hypotenuse = the slope

ehild
Homework Helper
No, h=10 m, and s is the hypotenuse. Look at the picture.

No, h=10 m, and s is the hypotenuse. Look at the picture.
ok so,

sin(θ)=(10/s)

ehild
Homework Helper
Isolate s and use it in the equation v2f - v2i = 2as.

so to isolate s, do you mean something like this?

s=10sin(θ)

ehild
Homework Helper
so to isolate s, do you mean something like this?

s=10sin(θ)
Something like that, but not quite. Is it more clear if I say solve the equation sin(θ)=(10/s) for s?

Something like that, but not quite. Is it more clear if I say solve the equation sin(θ)=(10/s) for s?
more like this?
s=10/sin(θ)

I'm sorry, this stuff has me so confused... These online classes aren't really meeting my expectations. :(

Last edited:
ehild
Homework Helper
Yes.
Determine the acceleration a now.

Yes.
Determine the acceleration a now.
well, acceleration = gsin(θ)
so, (9.81)sin(θ)=a

ehild
Homework Helper
Use a and s in the formula v2f - v2i = 2as.

well v2i = 0 since the object starts at rest
v2f = 2((9.81)sin(θ)) (10/sin(θ))

ehild
Homework Helper
Good! Evaluate.

I'm not sure if this is right, but it's a shot in the dark...

v2f =2 ((9.81)sin(θ) (10/sin(θ))
umm??? sin(θ) cancels out on each side leaving, 2(9.81)(10) = 196.2
v2f = 196.2
vf = √196.2 = 14m/s

is this even close?

ehild
Homework Helper
It is correct. That is the speed at the bottom of the slope when the whole length is covered. What is the speed, when the object is halfway down the slope?

It is correct. That is the speed at the bottom of the slope when the whole length is covered. What is the speed, when the object is halfway down the slope?
lol i would say half of that? Thats just me guessing...

ehild
Homework Helper
lol i would say half of that? Thats just me guessing...
NO!!!!!!!!!!!!! what is the new distance s(half) if it is half the previous one?

NO!!!!!!!!!!!!! what is the new distance s(half) if it is half the previous one?
haha I knew it wouldn't be that easy ;)

v2f = 2 * 9.81(10 * .5)
vf = √98.1
vf = 9.9m/s

ehild
Homework Helper
You did it!

• keithcuda
I feel the need to celebrate!!!
Thank you very much!

When I started this class, I really thought it would be one of my favorite subjects. I like it, but it really has me scratching my head.

I greatly appreciate it!
Merry Christmas!!

Cheers!

ehild
Homework Helper
Merry Christmas to you, too. :)