Question 211578
{{{h=-16t^2+400t}}} Start with the given equation.



{{{0=-16t^2+400t}}} Plug in {{{h=0}}} (since the ground has a height of 0 ft)



{{{-16t^2+400t=0}}} Combine like terms.



Notice that the quadratic {{{-16t^2+400t+0}}} is in the form of {{{At^2+Bt+C}}} where {{{A=-16}}}, {{{B=400}}}, and {{{C=0}}}



Let's use the quadratic formula to solve for "t":



{{{t = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{t = (-(400) +- sqrt( (400)^2-4(-16)(0) ))/(2(-16))}}} Plug in  {{{A=-16}}}, {{{B=400}}}, and {{{C=0}}}



{{{t = (-400 +- sqrt( 160000-4(-16)(0) ))/(2(-16))}}} Square {{{400}}} to get {{{160000}}}. 



{{{t = (-400 +- sqrt( 160000-0 ))/(2(-16))}}} Multiply {{{4(-16)(0)}}} to get {{{0}}}



{{{t = (-400 +- sqrt( 160000 ))/(2(-16))}}} Subtract {{{0}}} from {{{160000}}} to get {{{160000}}}



{{{t = (-400 +- sqrt( 160000 ))/(-32)}}} Multiply {{{2}}} and {{{-16}}} to get {{{-32}}}. 



{{{t = (-400 +- 400)/(-32)}}} Take the square root of {{{160000}}} to get {{{400}}}. 



{{{t = (-400 + 400)/(-32)}}} or {{{t = (-400 - 400)/(-32)}}} Break up the expression. 



{{{t = (0)/(-32)}}} or {{{t =  (-800)/(-32)}}} Combine like terms. 



{{{t = 0}}} or {{{t = 25}}} Reduce. 



So the solutions are {{{t = 0}}} or {{{t = 25}}}



Since we already know that the projectile is on the ground at t=0 seconds, this means that we can ignore this solution.



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Answer:


So the solution is {{{t=25}}} which means that it will take 25 seconds for the projectile to return to Earth.