Conservation of Energy initial velocity

[vesperaka]

I know if there was no initial velocity, the velocity at the bottom would just be sqrt(2gh), but I'm not sure how the initial velocity impacts the velocity at the bottom. I thought just adding the x-component of the velocity would be enough but apparently it's not :\

 

[chislam]

Well if there was an initial velocity, then what was the initial kinetic energy? Just remember:

$$KE_0 + PE_0 = KE + PE$$

 

[tiny-tim]

Welcome to PF!

I know if there was no initial velocity, the velocity at the bottom would just be sqrt(2gh), but I'm not sure how the initial velocity impacts the velocity at the bottom. I thought just adding the x-component of the velocity would be enough but apparently it's not :\

Hi vesperaka! Welcome to PF!

Yes, do what chislam suggests.

Another way of looking at it: your equation can be written (1/2)v₁² = gh.

So how would you put v₀ into that?

 

[vesperaka]

I understand that KE = 1/2mv^2, so when I tinker around with that I get

V = sqrt (2KE/m)

But that equation gives me a problem cause I can't just make KE a variable like that. I do know that all of the energy at the bottom is KE though. So:

1/2mv_i^2 + sqrt(2gh) = 1/2mv_f^2

When I take that equation and solve for Vf, I get this really long equation

Vf = V_i^2 + [2 sqrt(2gh)] / m


Is that the right answer? If I answer it incorrectly 1 more time I don't get credit so I'm being cautious, but to be honest I don't understand why it wouldn't work (unless I botched the algebra).

 

[tiny-tim]

… 1/2mv_i^2 + sqrt(2gh) = 1/2mv_f^2

When I take that equation and solve for Vf, I get this really long equation

Vf = V_i^2 + [2 sqrt(2gh)] / m

Hi vesperaka!

No, you're very confused.

KE = 1/2 mv^2. PE = mgh (not gh, and certainly not 2gh).

And where did that sqrt come from?

Try again!