# Using Acceleration to Calculate Gravity

1. Feb 6, 2010

### zx95

1. Two steel balls are suspended from electromagnets, and are released simultaneously when the electric current is shut off. They start at heights X1 and X2, with X2 being higher. When they hit the ground, they land T seconds apart. Find an equation for g (the gravitational field) in terms of T, X1 and X2

2. No further information was given.

3. I have about 8 pages of scratch paper and 5 hours of work trying to figure out what is probably something simple. I began by trying to figure out the acceleration from X1 to the ground:

0 = (1/2)at^2 + Vt + X1
-X1 = (1/2)at^2 - Vt
-X1 - Vt = (1/2)at^2
-2(X1+Vt)/t^2 = a

With this equation, I thought I'd be able to somehow figure out the velocity after the ball from X2 height falls from X2 to X1 and then plug that into my result above.

I tried many different things and none seem to match up with the answer my professor is looking for. A nudge in the right direction is all I'm looking for. After so much work, it's really frustrating to be at square 1 still. Thank you in advance for any light you can shed.

2. Feb 6, 2010

### rl.bhat

Hi zx95, welcome to PF.
Initial velocities of the steel balls are zero.
So xi = 1/2*g*t1^2. You can rewrite it as
sqrt(2x1/g) = t1.
Similarly find the expression for t2. Then find t2 - t1 which is equal to T.

3. Feb 6, 2010

### zx95

Thank you for the quick response!

I took this information and came up with the following:

t1 = sqrt(2X1/g) t2 = sqrt(2X2/g)

T = t2 - t1
T = sqrt(2X2/g) - sqrt (2X1/g)
T = sqrt(2(X2-X1)/g)
sqrt(g) = sqrt(2(X2-X1))/T

g = 2(X2-X1)/T^2

I plugged this into the professors software (and looking back in my scratch paper this is the same answer I got through a much more difficult process). Is this answer correct, or am I missing some step that I'm not seeing?

4. Feb 6, 2010

### rl.bhat

T = sqrt(2X2/g) - sqrt (2X1/g)
T = sqrt(2(X2-X1)/g)

The simplification is wrong.
It should be
T = [ sqrt(2x2) - sqrt(2x1)]/sqrt(g). so
sqrt(g) = [ sqrt(2x2) - sqrt(2x1)]/T

5. Feb 7, 2010

### zx95

Thank you very much! I always make mistakes like that. I need to work on that. Thanks!