Also, is this a standard equation or is it derived?
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u is initial velocity.
It is a derived one, by the law of energy conservation
=>KE(final) = KE(initial) + PE(initial)
=>1/2mv^2 = 1/2mu^2 + mgh
=>v^2 = u^2 + 2gh
It is a derived one, by the law of energy conservation
=>KE(final) = KE(initial) + PE(initial)
=>1/2mv^2 = 1/2mu^2 + mgh
=>v^2 = u^2 + 2gh
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There may be other instances where this equation would apply, but the one you're most likely to encounter is for a body moving with initial speed, u, in a uniform gravitational field of acceleration, g.
If it is not acted on by any forces other than gravity (or only by other forces that do no net work on it), and at some other time (whether it's earlier or later!) its height is lower by h, then its speed at that point will be v.
The equation can be derived by equating total energy, potential + kinetic, at the two different positions. Doing this makes the increase in kinetic energy equal to the decrease in potential energy, so that you have:
½mv² - ½mu² = mgh; "m" cancels out, and by rearranging, you get
v² = u² + 2gh
where:
u = speed at time 1
v = speed at time 2
h = (height at time 1) - (height at time 2)
g = acceleration of gravity
[Note that this works whether "h" is positive or negative!]
If it is not acted on by any forces other than gravity (or only by other forces that do no net work on it), and at some other time (whether it's earlier or later!) its height is lower by h, then its speed at that point will be v.
The equation can be derived by equating total energy, potential + kinetic, at the two different positions. Doing this makes the increase in kinetic energy equal to the decrease in potential energy, so that you have:
½mv² - ½mu² = mgh; "m" cancels out, and by rearranging, you get
v² = u² + 2gh
where:
u = speed at time 1
v = speed at time 2
h = (height at time 1) - (height at time 2)
g = acceleration of gravity
[Note that this works whether "h" is positive or negative!]