Question 696357
 Use the vertical height formula to solve the following h= -16t^ + initial velocity + initial height.
 A person throws a ball in the air with an initial height of 4 feet and an initial velocity of 40 feet per second. 
The person catches the ball when it is 3 feet from the ground.
 how much time is the ball in the air?
:
The equation for this situation:
-16t^2 + 40t + 4 = 3
-16t^2 + 40t + 4 - 3 = 0
-16t^2 + 40t + 1 = 0
Use the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}. where
x = t; a = -16; b = 40; c = 1
{{{t = (-40 +- sqrt(40^2-4*-16*1 ))/(2*-16) }}}
:
{{{t = (-40 +- sqrt(1600^2 + 64 ))/(-32) }}} 
:
{{{t = (-40 +- sqrt(1664 ))/(-32) }}}
Two solutions
{{{t = (-40 + 40.79)/(-32) }}} 
t = {{{(+.79)/(-32)}}}
t = -.024755; not a reasonable solution
and
{{{t = (-40 - 40.79)/(-32) }}} 
t = {{{(-80.79)/(-32)}}}
t = +2.52 seconds is the time it is in the air and caught