Question 176934
Since we have a "starting velocity of 10 ft/s from a height of 3 ft", this means that {{{v=10}}} and {{{c=3}}}. So the equation is {{{h= -16t^2+10t+3}}}


The ball will be in the air as long as the height h is NOT zero (when the ball is on the ground). So if we plug in {{{h=0}}} and solve for t, we'll find when the ball hits the ground. From there, we can figure out how long it will be in the air.




{{{h= -16t^2+10t+3}}} Start with the given equation.



{{{0= -16t^2+10t+3}}} Plug in {{{h=0}}}



Notice we have a quadratic equation in the form of {{{at^2+bt+c}}} where {{{a=-16}}}, {{{b=10}}}, and {{{c=3}}}



Let's use the quadratic formula to solve for t



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



{{{t = (-(10) +- sqrt( (10)^2-4(-16)(3) ))/(2(-16))}}} Plug in  {{{a=-16}}}, {{{b=10}}}, and {{{c=3}}}



{{{t = (-10 +- sqrt( 100-4(-16)(3) ))/(2(-16))}}} Square {{{10}}} to get {{{100}}}. 



{{{t = (-10 +- sqrt( 100--192 ))/(2(-16))}}} Multiply {{{4(-16)(3)}}} to get {{{-192}}}



{{{t = (-10 +- sqrt( 100+192 ))/(2(-16))}}} Rewrite {{{sqrt(100--192)}}} as {{{sqrt(100+192)}}}



{{{t = (-10 +- sqrt( 292 ))/(2(-16))}}} Add {{{100}}} to {{{192}}} to get {{{292}}}



{{{t = (-10 +- sqrt( 292 ))/(-32)}}} Multiply {{{2}}} and {{{-16}}} to get {{{-32}}}. 



{{{t = (-10 +- 2*sqrt(73))/(-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>)  



{{{t = (-10+2*sqrt(73))/(-32)}}} or {{{t = (-10-2*sqrt(73))/(-32)}}} Break up the expression.  



{{{t=-0.222}}} or {{{t=0.847}}} Use a calculator to approximate the answers.



Since a negative time value does NOT make sense, this means that the only solution is {{{t=0.847}}}



So when the time is approximately 0.847 seconds, the ball will hit the ground. 



So this means that the ball will be in the air for about 0.847 seconds.