Question 118286
Since we're finding the time it takes for the ball to reach 25 feet, this means h is 25


{{{h=-16t^2+40t}}} Start with the given equation



{{{25=-16t^2+40t}}} Plug in {{{h=25}}}



{{{16t^2-40t+25=0}}} Add {{{16t^2}}} to both sides. Subtract {{{40t}}} from both sides.



Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{16*x^2-40*x+25=0}}} ( notice {{{a=16}}}, {{{b=-40}}}, and {{{c=25}}})





{{{x = (--40 +- sqrt( (-40)^2-4*16*25 ))/(2*16)}}} Plug in a=16, b=-40, and c=25




{{{x = (40 +- sqrt( (-40)^2-4*16*25 ))/(2*16)}}} Negate -40 to get 40




{{{x = (40 +- sqrt( 1600-4*16*25 ))/(2*16)}}} Square -40 to get 1600  (note: remember when you square -40, you must square the negative as well. This is because {{{(-40)^2=-40*-40=1600}}}.)




{{{x = (40 +- sqrt( 1600+-1600 ))/(2*16)}}} Multiply {{{-4*25*16}}} to get {{{-1600}}}




{{{x = (40 +- sqrt( 0 ))/(2*16)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (40 +- 0)/(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>)




{{{x = (40 +- 0)/32}}} Multiply 2 and 16 to get 32


So now the expression breaks down into two parts


{{{x = (40 + 0)/32}}} or {{{x = (40 - 0)/32}}}


Lets look at the first part:


{{{x=(40 + 0)/32}}}


{{{x=40/32}}} Add the terms in the numerator

{{{x=5/4}}} Divide


So one answer is

{{{x=5/4}}}




Now lets look at the second part:


{{{x=(40 - 0)/32}}}


{{{x=40/32}}} Subtract the terms in the numerator

{{{x=5/4}}} Divide


So another answer is

{{{x=5/4}}}


So our only solution is:


{{{x=5/4}}} (which is {{{x=1.25}}} in decimal form)


So at 1.25 seconds, the ball is 25 feet high.