<PRE><B>Question 1040847</B>
A ball is thrown into the air with an initial velocity of 64 feet per second from a height of 192 feet.
 After how many seconds will the ball return to the ground and have a height of zero again?
:
Using the equation:
  t = time
 -16t^2 force of gravity downward,
 64t = initial velocity upward,
 192 = initial height 
:
-16t^2 + 64t + 192 = 0
simplify, divide equation by -16
t^2 - 4t - 12 = 0
Factors easily to 
(t-6)(t+2) = 0
the positive solution
t = 6 sec to return to the ground
:
:
see if that works
-16(6^2) + 64(6) + 192 =
-576 + 384 + 192 = 0

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