SOLUTION: When the baton twirler releases the baton into the air with an initial velocity of 30ft per sec and 5ft from the ground. She will catch the baton when it falls back to a height of

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Question 139239: When the baton twirler releases the baton into the air with an initial velocity of 30ft per sec and 5ft from the ground. She will catch the baton when it falls back to a height of 6ft, How many seconds is the baton in the air?

h=initial height
t=time in motion
s=initial height
v=initial velocity
Please and Thanks a bunch!
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
When the baton twirler releases the baton into the air with an initial velocity of 30ft per sec and 5ft from the ground. She will catch the baton when it falls back to a height of 6ft, How many seconds is the baton in the air?
:
h=-16t^2 + vt + s
:
h= final height (6 ft); I changed this to the "final height"
t= time in motion (solve for the time in seconds)
s= initial height 5 ft
v=initial velocity 30 ft/sec
:
-16t^2 + vt + s = h
:
Substitute the given values:
-16t^2 + 30t + 5 = 6
;
-16t^2 + 30t + 5 - 6 = 0
:
-16t^2 + 30t - 1 = 0
:
We need to use the quadratic formula here:

In this equation: a=-16; b=30; c=-1

The positive solution:

t = +1.84 sec is the time the baton is in the air
:
:
you can check the correctness of this solution, substitute 1.84 for t in:
h= -16t^2 + 30t + 5 to see if you get h~6