Question 102209
{{{h=-5t^2+20t+100}}} Start with the given equation (assuming you did everything right for problem 35)


When the height is 80, this means h=80


{{{80=-5t^2+20t+100}}} So plug in h=80




{{{0=-5t^2+20t+20}}} Subtract 80 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 {{{-5*t^2+20*t+20=0}}} ( notice {{{a=-5}}}, {{{b=20}}}, and {{{c=20}}})





{{{t = (-20 +- sqrt( (20)^2-4*-5*20 ))/(2*-5)}}} Plug in a=-5, b=20, and c=20




{{{t = (-20 +- sqrt( 400-4*-5*20 ))/(2*-5)}}} Square 20 to get 400  




{{{t = (-20 +- sqrt( 400+400 ))/(2*-5)}}} Multiply {{{-4*20*-5}}} to get {{{400}}}




{{{t = (-20 +- sqrt( 800 ))/(2*-5)}}} Combine like terms in the radicand (everything under the square root)




{{{t = (-20 +- 20*sqrt(2))/(2*-5)}}} 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 = (-20 +- 20*sqrt(2))/-10}}} Multiply 2 and -5 to get -10


So now the expression breaks down into two parts


{{{t = (-20 + 20*sqrt(2))/-10}}} or {{{t = (-20 - 20*sqrt(2))/-10}}}



Now break up the fraction



{{{t=-20/-10+20*sqrt(2)/-10}}} or {{{t=-20/-10-20*sqrt(2)/-10}}}



Simplify



{{{t=2-2*sqrt(2)}}} or {{{t=2+2*sqrt(2)}}}



So these expressions approximate to


{{{t=-0.82842712474619}}} or {{{t=4.82842712474619}}}



So our possible solutions are:

{{{t=-0.82842712474619}}} or {{{t=4.82842712474619}}}




However, since a negative time doesn't make sense, our only solution is {{{t=4.82842712474619}}}



So at about 4.8 seconds the ball will be 80 m high