1. Sep 30, 2007

### graten2go

Thank you for everyones help.

Last edited: Sep 30, 2007
2. Sep 30, 2007

### Sourabh N

Sorry but I didn't understand your solution. :( But my approach will be : Calculate distance traveled by them at any instant and add them up. There sum is equal to height of tower.

3. Sep 30, 2007

### johnsonandrew

We have two variables, time and height, and two equations, one for the first object and one for the second.

The equation for the first object is x= (xi) -(vi)(t) - (1/2)(g)(t^2) , where xi= initial height= 60 m
The equation for the second object, since g is working against the motion, is x= (xi) + (Vi)(t)-(1/2)(g)(t^2), vi= initial velocity, and xi= initial position/height= 0 m

Plug in your knowns, and solve the system of equations.

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On second look I see that you did this for your second attempt, and this is correct as far as I know-- 40.9 m. There must be a mistake in your first attempt.

Your first attempt is confusing, I can't really tell what you did. Maybe your position-time equation for the first object is incorrect. What equation did you use for your first object? Leave out units for now so I can see what you did.

Last edited: Sep 30, 2007