Questions on Algebra: Quadratic Equation answered by real tutors!

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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.

You can put this solution on YOUR website!
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?
:
The equation for this, s(t)= height, t=time in seconds
:
s(t) = -16t^2 + 30t + 5
:
Find the time that the baton is at 6 ft
:
-16t^2 + 30t + 5 = 6
:
-16t^2 + 30t + 5 - 6 = 0
:
-16t^2 + 30t - 1 = 0
;
use the quadratic formula to solve this

in this equation a=-16; b=30; c=-1

;

:

Two solutions:


t = +.034 sec (at 6' on the way up)
and


t = +1.84 sec (at 6' on the way down)