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?
:
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 = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
in this equation a=-16; b=30; c=-1
{{{t = (-30 +- sqrt(30^2- 4 *-16*-1 ))/(2*-16) }}}
;
{{{t = (-30 +- sqrt(900 - 64 ))/(-32) }}}
:
{{{t = (-30 +- sqrt(836))/(-32) }}}
Two solutions:
{{{t = (-30 + 28.91)/(-32) }}}
{{{t = (-1.09)/(-32) }}}
t = +.034 sec (at 6' on the way up)
and
{{{t = (-30 - 28.91)/(-32) }}}
{{{t = (-58.91)/(-32) }}}
t = +1.84 sec (at 6' on the way down)