Question 90295
{{{h = -16t^2 + 112t }}}


{{{180 = -16t^2 + 112t }}} Let h=180


{{{0 = -16t^2 + 112t -180}}} Subtract 180 from both sides



Let's use the quadratic formula to solve for t:



Starting with the general quadratic


{{{at^2+bt+c=0}}}


the general solution using the quadratic equation is:


{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{-16*t^2+112*t-180=0}}} ( notice {{{a=-16}}}, {{{b=112}}}, and {{{c=-180}}})


{{{t = (-112 +- sqrt( (112)^2-4*-16*-180 ))/(2*-16)}}} Plug in a=-16, b=112, and c=-180




{{{t = (-112 +- sqrt( 12544-4*-16*-180 ))/(2*-16)}}} Square 112 to get 12544  




{{{t = (-112 +- sqrt( 12544+-11520 ))/(2*-16)}}} Multiply {{{-4*-180*-16}}} to get {{{-11520}}}




{{{t = (-112 +- sqrt( 1024 ))/(2*-16)}}} Combine like terms in the radicand (everything under the square root)




{{{t = (-112 +- 32)/(2*-16)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{t = (-112 +- 32)/-32}}} Multiply 2 and -16 to get -32


So now the expression breaks down into two parts


{{{t = (-112 + 32)/-32}}} or {{{t = (-112 - 32)/-32}}}


Lets look at the first part:


{{{x=(-112 + 32)/-32}}}


{{{t=-80/-32}}} Add the terms in the numerator

{{{t=5/2}}} Divide


So one answer is

{{{t=5/2}}}




Now lets look at the second part:


{{{x=(-112 - 32)/-32}}}


{{{t=-144/-32}}} Subtract the terms in the numerator

{{{t=9/2}}} Divide


So another answer is

{{{t=9/2}}}


So our solutions are:

{{{t=5/2}}} or {{{t=9/2}}}

which are 


{{{t=2.5}}} or {{{t=4.5}}}


in decimal form


So it takes 2.5 seconds to rise to 180 feet. Then it takes 4.5 seconds (elapsed from the beginning time) to fall back to 180 feet.