time for an object to fall


by jamesd2008
Tags: fall, object, time
jamesd2008
jamesd2008 is offline
#1
Apr25-10, 06:06 PM
P: 64
Hi,

If you no the height of an object only, assuming no frictional forces, and that it is on earh. Can you determine how long it will take to hit the ground?

Thanks
James
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jamesd2008
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#2
Apr25-10, 06:08 PM
P: 64
Also the final velocity is also unknown and the initial velocity is zero?
karkas
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#3
Apr25-10, 07:45 PM
P: 125
I am assuming that you're talking about the height between the object and the floor. The answer is yes, and that is obvious from the equations that describe free fall, meaning the fall that is induced and preserved only by the force of gravitational attraction that the earth exerts on the body.

So we have
[latex]h = \frac{1}{2} g t^2 \Leftrightarrow t = \sqrt{\frac{2h}{g}} [/latex]
and since you know h, you can calculate the required time. Using that t you can solve for the final velocity of the body, the one it has right before it reaches ground. And yes, since you let the body go at one point without pushing it downwards, the initial velocity is zero.

morrobay
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#4
Apr25-10, 09:53 PM
PF Gold
P: 358

time for an object to fall


Quote Quote by karkas View Post
I am assuming that you're talking about the height between the object and the floor. The answer is yes, and that is obvious from the equations that describe free fall, meaning the fall that is induced and preserved only by the force of gravitational attraction that the earth exerts on the body.

So we have
[latex]h = \frac{1}{2} g t^2 \Leftrightarrow t = \sqrt{2gh} [/latex]
and since you know h, you can calculate the required time. Using that t you can solve for the final velocity of the body, the one it has right before it reaches ground. And yes, since you let the body go at one point without pushing it downwards, the initial velocity is zero.
No t=square root (2h/g)
velocity = square root (2gh)
karkas
karkas is offline
#5
Apr26-10, 06:40 AM
P: 125
Yes sorry my bad.
jamesd2008
jamesd2008 is offline
#6
Apr26-10, 09:49 AM
P: 64
Thanks for the reply's guys, much help


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