SOLUTION: A ball is thrown straight upward from the top of a building 50 feet tall at an initial velocity of 100 feet per second. The equation s=-16^2+100t+50 gives teh heights s of the bal

Question 379761: A ball is thrown straight upward from the top of a building 50 feet tall at an initial velocity of 100 feet per second. The equation s=-16^2+100t+50 gives teh heights s of the ball t seconds afer it is thrown. After how many seconds does the ball reach its maximum height?
I started with -100/2(-16) but not sure that is even where to start.
Thank you for your help
Tracy
Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website!
You were okay.
The key to doing this is to realize that when the ball reaches its maximum height, its velocity is ZERO. We can then find the time it's in the air.
We know v(f) - v(i) = gt. Here v(final) is zero and v(initial) is 100. g = -32.
Therefore -100 = -32t and t = 3.125 seconds.

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