Question 136849
Desperate, in order to do this problem, you need the formula for distance of a dropped object.

I'll assume your math or physics book has that in it. The general equation is
{{{d = -(a/2)t^2 + v[0]t + d[0]}}} where a is gravity, v{0] is initial velocity and d[0] is initial distance.

In your case, the ball is 'just dropped'. So there is no initial velocity --> v[0] = 0.

d[0] is given at 60 feet (funny when you think that Galileo was a metric kind of guy).

So we have {{{d = -(a/2)t^2 + 60 }}}

What you need to 'know' is that gravity measures 32ft/sec^2.

{{{d = -16t^2 + 60 }}}

What is d when the objects hit the ground? d = 0

{{{0 = -16t^2 + 60}}}
{{{-60 = -16t^2}}}
{{{15/4 = t^2}}}

You do the rest :-D