Question 466060
The height of an object thrown upward from the floor of a canyon 106 ft. deep,
 with an initial velocity of 120 ft/sec is given by the equation:
h=-16t2 + 120t-106 How long will it take the object to rise to the height of the canyon wall?
 Round answer to the nearest hundreth.
:
Height of the canyon wall = 0, depth of the canyon -106, so we have
-16t^2 + 120t - 106 = 0
Use the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this equation: t=x, a=-16, b-120, c=-106 
{{{t = (-120 +- sqrt(120^2-4*-16*-106 ))/(2*-16) }}}
:
{{{t = (-120 +- sqrt(14400-6784 ))/(-32) }}}
:
{{{t = (-120 +- sqrt(7616 ))/(-32) }}}
Two solutions
{{{t = (-120 + 87.27)/(-32) }}}
t = {{{(-32.73)/(-32)}}}
t = 1.023 sec, this is the solution we want, time at the top of the wall on the way up