# How do I finish this off?

1. Sep 28, 2009

### neutron star

I think I'm close but I don't know what the next step is.

1. The problem statement, all variables and given/known data
A ball is dropped off a cliff on a planet with an exact gravity of $$10m/s^2$$, 1.5 seconds later a lead ball is thrown straight down with an initial speed of $$20 m/s$$ If the two balls hit the base of the cliff at the same time, find the height of the cliff.

2. Relevant equations
$$X$$f=$$X$$o+$$V$$o$$t$$ -$$1/2gt^2$$

3. The attempt at a solution$$-d=-1/2gt^2$$
$$t^2=2d/g$$
$$t=\sqrt{2d/g}$$

$$-d=0-20(\sqrt{2d/g}-1.5)-1/2g(\sqrt{2d/g}-1.5)^2$$

Last edited: Sep 28, 2009
2. Sep 28, 2009

### Delphi51

Wow, it took me a while to see that your last expression is d = Vi*t + .5at^2 with t replaced by (t-1.5). So that is your distance for the second ball at time t from when the first ball is dropped. To finish the job, you equate that distance to a similar expression for the first ball's distance.

If I may offer a more general tip, it would be to begin with
ball 1 distance = ball 2 distance
.5*a*t^2 = Vi*(t-1.5) + .5a(t-1.5)^2
so you have the big picture clearly stated, and then it is just detail work to finish it.
Also, you might take "down" to be positive in this problem to reduce the number of troublesome minus signs. I see it isn't really a quadratic equation - the t^2 term cancels a step or two further on.