Question 124287
The object will hit the ground when h=0



{{{h = -16t^2 + 30t + 3}}} Start with the given equation



{{{0 = -16t^2 + 30t + 3}}} 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+30*t+3=0}}} ( notice {{{a=-16}}}, {{{b=30}}}, and {{{c=3}}})





{{{t = (-30 +- sqrt( (30)^2-4*-16*3 ))/(2*-16)}}} Plug in a=-16, b=30, and c=3




{{{t = (-30 +- sqrt( 900-4*-16*3 ))/(2*-16)}}} Square 30 to get 900  




{{{t = (-30 +- sqrt( 900+192 ))/(2*-16)}}} Multiply {{{-4*3*-16}}} to get {{{192}}}




{{{t = (-30 +- sqrt( 1092 ))/(2*-16)}}} Combine like terms in the radicand (everything under the square root)




{{{t = (-30 +- 2*sqrt(273))/(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 = (-30 +- 2*sqrt(273))/-32}}} Multiply 2 and -16 to get -32


So now the expression breaks down into two parts


{{{t = (-30 + 2*sqrt(273))/-32}}} or {{{t = (-30 - 2*sqrt(273))/-32}}}



Now break up the fraction



{{{t=-30/-32+2*sqrt(273)/-32}}} or {{{t=-30/-32-2*sqrt(273)/-32}}}



Simplify



{{{t=15 / 16-sqrt(273)/16}}} or {{{t=15 / 16+sqrt(273)/16}}}



So these expressions approximate to


{{{t=-0.095169477616144}}} or {{{t=1.97016947761614}}}



So our possible solutions are:

{{{t=-0.095169477616144}}} or {{{t=1.97016947761614}}}



However, since a negative time doesn't make sense, our only solution is {{{t=1.97016947761614}}}



So the object will hit the ground at about 1.970 seconds