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

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> 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      Log On


   



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) About Me  (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) About Me  (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
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
in this equation a=-16; b=30; c=-1
t+=+%28-30+%2B-+sqrt%2830%5E2-+4+%2A-16%2A-1+%29%29%2F%282%2A-16%29+
;
t+=+%28-30+%2B-+sqrt%28900+-+64+%29%29%2F%28-32%29+
:
t+=+%28-30+%2B-+sqrt%28836%29%29%2F%28-32%29+
Two solutions:
t+=+%28-30+%2B+28.91%29%2F%28-32%29+
t+=+%28-1.09%29%2F%28-32%29+
t = +.034 sec (at 6' on the way up)
and
t+=+%28-30+-+28.91%29%2F%28-32%29+
t+=+%28-58.91%29%2F%28-32%29+
t = +1.84 sec (at 6' on the way down)