its the acceleration due to gravity on Earth in feet per second per second

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- Jul 6th 2009, 09:55 PM #1

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## Y(T) = -(1/2)GT^2 + VoT + Yo?

A stone is dropped from a height of 64 feet. In how many seconds does it hit the ground? What is the speed at impact?

I don't think you guys can see the formula, but I want to know where did the -32 come from? If its suppose to be cause of -(1/2)64 then why is there still a (1/2) there and why does that (1/2) turn it into a -16?

My just don't understand cause the formula I was told is.

Y(T) = -(1/2)GT^2 + VoT + Yo.

- Jul 6th 2009, 09:58 PM #2

- Jul 6th 2009, 10:07 PM #3

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- Jul 6th 2009, 10:14 PM #4

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- Jul 6th 2009, 10:21 PM #5

- Jul 6th 2009, 10:24 PM #6
You have here the fundamental equation representative of a freely falling object.

Recall that or . So the acceleraton of a freely falling object (first measured by galileo) is the constant . The is a factor of because we must take the average from to because we must interpret as a distance. In your formula, we have to also consider the initial velocity of the object because that will add to the velocity at any time and thereby affect distance traveled. And of course, since we are concerned with distance, we consider the initial height. Note that at , the object is still at rest, at it's initial height.

I hope this helps you to conceptualize what is happening here.

- Jul 6th 2009, 10:33 PM #7

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In fact

G is a constant ( Umm i don't know what the value is )

M is the mass of the Earth .

R is the distance from the center of the Earth ( about 6300 km)

Since these values are extremely large or small , g won't be affected if R increases or decrases slightly .

- Jul 6th 2009, 10:46 PM #8

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- Jul 7th 2009, 12:08 AM #9
I agree with Caaptain Black--When I was a student in the early 80's we were convinced our children would only know of feet and inches in ancient literature.

However, I would like to comment on Von Nemo's Post. There is no averaging involved itis a matter of antidifferentiation.

We start with acceleration a = -32

By integration we get v = -gt + C

where v(0) = V0 hence v = -gt + V0 for the velocity

For position we integrate velocity y(t) = -1/2gt^2 +V0t +C

This is where the 1/2 comes in

again applying the initial condition y(t) = -1/2 gt^2 +V0t + Y0 for position

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