Question 122530
When the ball hits the ground, this means {{{h(t)=0}}} (in other words, the height is zero)


{{{h(t) = -16t^2 + 64t + 80 }}} Start with the given equation



{{{0= -16t^2 + 64t + 80 }}} Plug in {{{h(t)=0}}}. In other words, replace h(t) with 0



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+64*t+80=0}}} ( notice {{{a=-16}}}, {{{b=64}}}, and {{{c=80}}})





{{{t = (-64 +- sqrt( (64)^2-4*-16*80 ))/(2*-16)}}} Plug in a=-16, b=64, and c=80




{{{t = (-64 +- sqrt( 4096-4*-16*80 ))/(2*-16)}}} Square 64 to get 4096  




{{{t = (-64 +- sqrt( 4096+5120 ))/(2*-16)}}} Multiply {{{-4*80*-16}}} to get {{{5120}}}




{{{t = (-64 +- sqrt( 9216 ))/(2*-16)}}} Combine like terms in the radicand (everything under the square root)




{{{t = (-64 +- 96)/(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 = (-64 +- 96)/-32}}} Multiply 2 and -16 to get -32


So now the expression breaks down into two parts


{{{t = (-64 + 96)/-32}}} or {{{t = (-64 - 96)/-32}}}


Lets look at the first part:


{{{x=(-64 + 96)/-32}}}


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

{{{t=-1}}} Divide


So one answer is

{{{t=-1}}}




Now lets look at the second part:


{{{x=(-64 - 96)/-32}}}


{{{t=-160/-32}}} Subtract the terms in the numerator

{{{t=5}}} Divide


So another answer is

{{{t=5}}}


So our solutions are:

{{{t=-1}}} or {{{t=5}}}




However, since a negative time doesn't make sense, our only solution is 


{{{t=5}}}




-----------------------------


Answer:



So at 5 seconds, the ball will hit the ground.