How to calculate the time it takes for an object to go from its ...

[xclash]

I totally don't understand how to calculate this. Hint: Use potential energy calculation (PEgravitational=mgh) m=mass, g=gravitational field pull (9. and h=height, estimated mass, and kinetic energy formula (KE=.5mv^2). I'll try figuring this out again, but it's a problem I've been stuck on! Thank you in advance!

[frankwu]

OK:

At its maximum height, assume the object is not moving up or down.  The object posseses

(1)  PE = mgh

When it hits the ground, it will have lost all its PE, which will have been converted into KE:

(2)  KE = .5*mv^2

The force acting on the object as it falls is

(3)  F = ma, where the "a" represents acceleration imparted to m when the force F on it is converted into motion.  In the context of the earth's gravity, the "a" is what we symbolize by g:

(4)  F = mg,  where

 g  = 9.8 m/s^2 = 9.8 (m/s)/s, (meters per second) per second.

Velocity is meters/second, and acceleration is the rate of change in velocity, or velocity/second.

If you know the acceleration, then you know the velocity after t seconds:

(5)  v = gt

Putting all of the above together:

(6)  mgh = .5*mv^2      since KE = PE at impact.  Substituting for v from (5), we get

(7)  mgh = .5 * m * (gt)^2  = .5 * m * g^2 * t^2  

Canceling terms common to each side, we get:

h = .5 * g * t^2

Rearranging and solving for t, we get

(  t = ?(2h / g)   <<---ANSWER

Notice that the time is not dependent on the mass.
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