Question 93468

To find out when the projectile will hit the ground, let h=0 and solve for t


{{{0=-16t^2+204t+28}}} Plug in h=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+204*t+28=0}}} ( notice {{{a=-16}}}, {{{b=204}}}, and {{{c=28}}})


{{{t = (-204 +- sqrt( (204)^2-4*-16*28 ))/(2*-16)}}} Plug in a=-16, b=204, and c=28




{{{t = (-204 +- sqrt( 41616-4*-16*28 ))/(2*-16)}}} Square 204 to get 41616  




{{{t = (-204 +- sqrt( 41616+1792 ))/(2*-16)}}} Multiply {{{-4*28*-16}}} to get {{{1792}}}




{{{t = (-204 +- sqrt( 43408 ))/(2*-16)}}} Combine like terms in the radicand (everything under the square root)




{{{t = (-204 +- 4*sqrt(2713))/(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 = (-204 +- 4*sqrt(2713))/-32}}} Multiply 2 and -16 to get -32


So now the expression breaks down into two parts


{{{t = (-204 + 4*sqrt(2713))/-32}}} or {{{t = (-204 - 4*sqrt(2713))/-32}}}



Now break up the fraction



{{{t=-204/-32+4*sqrt(2713)/-32}}} or {{{t=-204/-32-4*sqrt(2713)/-32}}}



Simplify



{{{t=51 / 8-sqrt(2713)/8}}} or {{{t=51 / 8+sqrt(2713)/8}}}



So these expressions approximate to


{{{t=-0.135808321552709}}} or {{{t=12.8858083215527}}}



So our possible solutions are:

{{{t=-0.135808321552709}}} or {{{t=12.8858083215527}}}



Since a negative time doesn't make sense, our only solution is  {{{t=12.8858083215527}}}


So the projectile will hit the ground at about 12.89 seconds