Originally Posted by **LittleWing** Yes I know my physics professor is very vague with his questions..thank you for answering and decyphering my cryptic question.
Here is another attempt at the second question....
Ball is dropped from a building and it takes twice as long for the ball to drop off the last 500m as it does for the ball to drop the first 'h' meters. Determine H..... T(2)= T(1) x 2 |

I would do this in a few steps (for the sake of clarity.)

Let the height of the building be H. When the ball is at 500 m, let this be at time T. Then we know that

y = y0 + v0*t + (1/2)at^2 gives:

500 = H - (1/2)*gT^2

We need another equation in H and T to solve this. So let's take the second part. The initial time is T and the final time is T + 2T = 3T. The ball starts at y0 = 500 m. So we have

y = y0 + v0*(t - t0) + (1/2)a(t - t0)^2

(This equation is the same as y = y0 + v0*t + (1/2)at^2 with the exception of a "displaced" time variable.)

So

0 = 500 + v0*(2T) - (1/2)g*(2T)^2

Notice that the v0 in this equation is equal to the final v in the first part. So we need v from the first part:

v = v0 + at

v = -gT

Putting this v into the equation for the second part of the fall we get as v0:

0 = 500 + (-gT)*(2T) - (1/2)g*(2T)^2

This is an equation for T. From here we can get H:

500 = H - (1/2)gT^2

and then h:

H = h + 500

-Dan