When Mongol drops his ball off the Hasselbluff Mountain Viewpoint, its height above the ground at time t is given by h = 64 - 1/2gt2. How long will the ball take to hit the ground? How fast is the ball going the instant before it hits the ground?
I found the answer to the first question to be: t = 8√2/√g
The derivative is v(t) = - gt
The Attempt at a Solution
v(t) = [lim t-->0] - g(8√2/√g + ∆t) - - g(8√2/√g)
v(t) = [lim t-->0] - g8√2/√g + g(8√2/√g)
v(t) = [lim t-->0] 0
I know that I'm doing something very obviously wrong, I just can't figure out what.