Question 88736
If we set the equation equal to zero, we can find out when the ball will hit the ground

{{{-16t^2+80t+96=0}}}


So 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+80*t+96=0}}} ( notice {{{a=-16}}}, {{{b=80}}}, and {{{c=96}}})


{{{t = (-80 +- sqrt( (80)^2-4*-16*96 ))/(2*-16)}}} Plug in a=-16, b=80, and c=96




{{{t = (-80 +- sqrt( 6400-4*-16*96 ))/(2*-16)}}} Square 80 to get 6400




{{{t = (-80 +- sqrt( 6400+6144 ))/(2*-16)}}} Multiply {{{-4*96*-16}}} to get {{{6144}}}




{{{t = (-80 +- sqrt( 12544 ))/(2*-16)}}} Combine like terms in the radicand (everything under the square root)




{{{t = (-80 +- 112)/(2*-16)}}} Simplify the square root




{{{t = (-80 +- 112)/-32}}} Multiply 2 and -16 to get -32


So now the expression breaks down into two parts


{{{t = (-80 + 112)/-32}}} or {{{t = (-80 - 112)/-32}}}


Lets look at the first part:


{{{x=(-80 + 112)/-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=(-80 - 112)/-32}}}


{{{t=-192/-32}}} Subtract the terms in the numerator

{{{t=6}}} Divide


So another answer is

{{{t=6}}}


So our solutions are:

{{{t=-1}}} or {{{t=6}}}


Notice when we graph {{{-16*x^2+80*x+96}}} (just replace t with x), we get:


{{{ graph( 500, 500, -11, 16, -11, 16,-16*x^2+80*x+96) }}}


and we can see that the roots are {{{t=-1}}} and {{{t=6}}}. This verifies our answer


Since a negative time doesn't make sense, so our only solution is {{{t=6}}}. So it will take 6 seconds for the ball to hit the ground.