Energy consv.

[chiddler]

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If a projectile is launched at some angle to the horizontal and I want to find out the vertical velocity, can I use v = sqrt(2gh)?

I'd think it's not allowed because that v will find absolute velocity, not a component thereof. But I think i'm wrong because EK used it.

Why is it allowed?

 

[dmf2682]

If a projectile is launched at some angle to the horizontal and I want to find out the vertical velocity, can I use v = sqrt(2gh)?

I'd think it's not allowed because that v will find absolute velocity, not a component thereof. But I think i'm wrong because EK used it.

Why is it allowed?

You can use it because that equation comes from equating potential and kinetic energy.

Mgh = .5 mv ^2

solve for v and there's your equation. Since potential energy is only in the vertical, you're only going to find vertical.

 

[chiddler]

But the maximum height changes based on angle launched. You should at least use v*sin(theta), no?

 

[dmf2682]

But the maximum height changes based on angle launched. You should at least use v*sin(theta), no?

Exactly.

 

[chiddler]

I don't understand. Are you agreeing with me or agreeing with EK? EK didn't use sin.

The question.

They solved it by starting with v=sqrt(2gh) then converting g into acceleration for the charged plate (g = a = F/m and F = qE)

 

[milski]

You don't really care about the horizontal motion in this question - it works out the same with any horizontal velocity, including 0. If you consider the motion in vertical direction, at the top the object has gained some potential energy (mgh) and lost all of the energy due to the vertical component of velocity (m*vy^2/2). Set them equal and you're done.

 

[chiddler]

You don't really care about the horizontal motion in this question - it works out the same with any horizontal velocity, including 0. If you consider the motion in vertical direction, at the top the object has gained some potential energy (mgh) and lost all of the energy due to the vertical component of velocity (m*vy^2/2). Set them equal and you're done.

oh that makes a lot of sense.

thanks very much.