v=sqrt2gh

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MedPR

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  1. Pre-Podiatry
Can you use this equation to find the final velocity in every situation where the following is true?

free fall from a height h
vo = 0
constant acceleration
 
yeah i think so! i should say if Vo in the y direction is zero... if you throw horizontally down then it should still apply
 
use linear motion formulas for free fall

What equation would be better to find the vmax of a free falling object?

Also, I was asking because I just applied v=sqrt2gh to a pendulum question. The pendulum bob was raised to a height of 5m above the ground then dropped. The question was what would be the approximate speed of the bob as it passed through equilibrium? For the pendulum bob, equilibrium is the point where v is at a maximum right?
 
What equation would be better to find the vmax of a free falling object?

Also, I was asking because I just applied v=sqrt2gh to a pendulum question. The pendulum bob was raised to a height of 5m above the ground then dropped. The question was what would be the approximate speed of the bob as it passed through equilibrium? For the pendulum bob, equilibrium is the point where v is at a maximum right?
A pendulum is not really a free falling object. Free falling means that only the gravity is acting on it, for the pendulum you also have tension from the string on which it's hanging.

With that said, your formula is still correct and is the easy way to solve the problem. You can apply it in any case where an amount of potential energy is converted to kinetic. For example, I could probably think of some sort of contraption with a pendulum lying on a table and being pushed by a compressed spring - you'll end up with a similar formula for the final speed of the pendulum. Or a better example - the release speed of a ball in a pinball.
 
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good question, i havn't seen that one before, since we usually care about freq
but here the acceleratoin wouldn't be g... it's gsin of angle... so it's small at first, but it's increasing to g (when it's at the lowest point)
EDIT: read somewhere
v = √{2gL[1-cos(a)]}
 
good question, i havn't seen that one before, since we usually care about freq
but here the acceleratoin wouldn't be g... it's gsin of angle... so it's small at first, but it's increasing to g
The acceleration does not really matter - the formula comes from mv^2/2=mgh. The potential energy was mgh, all of it was converted to kinetic, which is mv^2/2. Solve for v and you're done.
 
i see your point but at the bottom the tension is higher too vs. when it's at an angle, so some of the energy should account for that?!

never mind, i'm confusing the formula they put which L-(blah), which is the h and i'm mixing this up with centripetal force, etc!
 
i see your point but at the bottom the tension is higher too vs. when it's at an angle, so some of the energy should account for that?!
The string is inelastic (normally), so while the tension is higher, there is no change in potential energy. Potential energy would be 1/2kx^2 where x is 0, so 0. Btw, the acceleration at the equilibrium point is also 0.
 
I see your point, I just confused myself with jumbo mumbo, thanks
 
A pendulum is not really a free falling object. Free falling means that only the gravity is acting on it, for the pendulum you also have tension from the string on which it's hanging.

With that said, your formula is still correct and is the easy way to solve the problem. You can apply it in any case where an amount of potential energy is converted to kinetic. For example, I could probably think of some sort of contraption with a pendulum lying on a table and being pushed by a compressed spring - you'll end up with a similar formula for the final speed of the pendulum. Or a better example - the release speed of a ball in a pinball.


Cool, thanks for clarifying that.

I tend to want to use it whenever I see a question involving "raised to a height" and "what is the velocity after being dropped"
 
What equation would be better to find the vmax of a free falling object?

Not sure if someone answered this since I didn't read the thread too carefully, but I think you can just first use X= X0 + V0t + 1/2at^2
which will give you the time it takes to fall
then use this equation : V = V0 + at

also now that I look at it...you can just use the one in your original post -_- well I put the effort into writing this now so I'm not gonna erase it.. 🙁
 
for pendulums you have to use KE=PE since that's what's changing all the time


Can you show me the setup for this particular problem?

You raise a pendulum bob 5m above the ground. What is the velocity when it passes through equilibrium?
 
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