SOLUTION: 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

Click here to see ALL problems on Quadratic Equations

Question 151131: 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?
Found 2 solutions by stanbon, ankor@dixie-net.com:
Answer by stanbon(75887) (Show Source):
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?
-------------------------------
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)
===============
Cheers,
Stan H.

Answer by ankor@dixie-net.com(22740) (Show Source):
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)