<PRE><B>Question 151131</B>
The baton twirler releases the baton into the air when it is 5 feet above the ground. The initial velocity of the baton is 30 feet per second. The twirler will catch the baton when it falls back to a height of 6 feet. How many seconds is the baton in the air?
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h(t) = -32t^2 + vot + so
6 = -32t^2 + 30t + 5
32t^2  - 30t + 1 = 0

t = [30 +- sqrt(30^2-4*32)]/64

t = [30 +- sqrt(772)]/64

t = [30 +- 27.785]/64

t = [0.0346 seconds (time the baton is at 6 ft. on the way up)
 or t = 0.9029 seconds (time the baton is at 6 ft. on the way down)
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Cheers,
Stan H.</PRE>