Question 195718
At what time will the ball reach its maximum height?
What is the maximum height of the ball?


First, we need the vertex. The vertex will tell us what the max height will be:



In order to find the vertex, we first need to find the x-coordinate of the vertex.



To find the x-coordinate of the vertex, use this formula: {{{x=(-b)/(2a)}}}.



{{{x=(-b)/(2a)}}} Start with the given formula.



From {{{y=-16x^2+80x+3}}}, we can see that {{{a=-16}}}, {{{b=80}}}, and {{{c=3}}}.



{{{x=(-(80))/(2(-16))}}} Plug in {{{a=-16}}} and {{{b=80}}}.



{{{x=(-80)/(-32)}}} Multiply 2 and {{{-16}}} to get {{{-32}}}.



{{{x=5/2}}} Reduce.



So the x-coordinate of the vertex is {{{x=5/2}}}. Note: this means that the axis of symmetry is also {{{x=5/2}}}.



Now that we know the x-coordinate of the vertex, we can use it to find the y-coordinate of the vertex.



{{{y=-16x^2+80x+3}}} Start with the given equation.



{{{y=-16(5/2)^2+80(5/2)+3}}} Plug in {{{x=5/2}}}.



{{{y=-16(25/4)+80(5/2)+3}}} Square {{{5/2}}} to get {{{25/4}}}.



{{{y=-100+80(5/2)+3}}} Multiply {{{-16}}} and {{{25/4}}} to get {{{-100}}}.



{{{y=-100+200+3}}} Multiply {{{80}}} and {{{5/2}}} to get {{{200}}}.



{{{y=103}}} Combine like terms.



So the y-coordinate of the vertex is {{{y=103}}}.



So the vertex is *[Tex \LARGE \left(\frac{5}{2},103\right)].



which in decimal form is *[Tex \LARGE \left(2.5,103\right)].



So this means that at 2.5 seconds, the ball will be at the max height of 103.



<hr>



When will the ball hit the ground?



{{{y= -16x^2 + 80x + 3}}} Start with the given equation.



{{{0= -16x^2 + 80x + 3}}} Plug in {{{y=0}}}



Notice we have a quadratic equation in the form of {{{0=ax^2+bx+c}}} where {{{a=-16}}}, {{{b=80}}}, and {{{c=3}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(80) +- sqrt( (80)^2-4(-16)(3) ))/(2(-16))}}} Plug in  {{{a=-16}}}, {{{b=80}}}, and {{{c=3}}}



{{{x = (-80 +- sqrt( 6400-4(-16)(3) ))/(2(-16))}}} Square {{{80}}} to get {{{6400}}}. 



{{{x = (-80 +- sqrt( 6400--192 ))/(2(-16))}}} Multiply {{{4(-16)(3)}}} to get {{{-192}}}



{{{x = (-80 +- sqrt( 6400+192 ))/(2(-16))}}} Rewrite {{{sqrt(6400--192)}}} as {{{sqrt(6400+192)}}}



{{{x = (-80 +- sqrt( 6592 ))/(2(-16))}}} Add {{{6400}}} to {{{192}}} to get {{{6592}}}



{{{x = (-80 +- sqrt( 6592 ))/(-32)}}} Multiply {{{2}}} and {{{-16}}} to get {{{-32}}}. 



{{{x = (-80 +- 8*sqrt(103))/(-32)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (-80+8*sqrt(103))/(-32)}}} or {{{x = (-80-8*sqrt(103))/(-32)}}} Break up the expression.  



So the <i>possible</i> answers are {{{x = (-80+8*sqrt(103))/(-32)}}} or {{{x = (-80-8*sqrt(103))/(-32)}}} 



which approximate to {{{x=-0.037}}} or {{{x=5.037}}} 



Since a negative time doesn't make sense, this means that the only solution is {{{x=5.037}}} 



So it will take about 5.037 seconds for the ball to hit the ground.