Question 211170
{{{h=vt-16t^2}}} Start with the given formula



{{{h=48t-16t^2}}} Plug in the given velocity of {{{v=48}}}



{{{0=48t-16t^2}}} Plug in the height of 0 {{{h=0}}} (since the height is zero at the ground)



{{{0=-16t^2+48t}}} Rearrange the terms.



Notice that the quadratic {{{-16t^2+48t}}} is in the form of {{{At^2+Bt+C}}} where {{{A=-16}}}, {{{B=48}}}, 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 = (-(48) +- sqrt( (48)^2-4(-16)(0) ))/(2(-16))}}} Plug in  {{{A=-16}}}, {{{B=48}}}, and {{{C=0}}}



{{{t = (-48 +- sqrt( 2304-4(-16)(0) ))/(2(-16))}}} Square {{{48}}} to get {{{2304}}}. 



{{{t = (-48 +- sqrt( 2304-0 ))/(2(-16))}}} Multiply {{{4(-16)(0)}}} to get {{{0}}}



{{{t = (-48 +- sqrt( 2304 ))/(2(-16))}}} Subtract {{{0}}} from {{{2304}}} to get {{{2304}}}



{{{t = (-48 +- sqrt( 2304 ))/(-32)}}} Multiply {{{2}}} and {{{-16}}} to get {{{-32}}}. 



{{{t = (-48 +- 48)/(-32)}}} Take the square root of {{{2304}}} to get {{{48}}}. 



{{{t = (-48 + 48)/(-32)}}} or {{{t = (-48 - 48)/(-32)}}} Break up the expression. 



{{{t = (0)/(-32)}}} or {{{t =  (-96)/(-32)}}} Combine like terms. 



{{{t = 0}}} or {{{t = 3}}} Simplify. 



So the solutions are {{{t = 0}}} or {{{t = 3}}} 

  

This means that at times of 0 and 3 seconds, the ball will be at the ground. 



Since we already know that the ball is at the ground at 0 seconds, the other solution {{{t=3}}} is more interesting. So we're going to ignore {{{t=0}}}



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Answer:



So it will take 3 seconds for the ball to return to the ground.