Question 490857
Initial velocity = 20
h(t) = -4.9t^2 +20t
Solve for t when h(t) = 5
--> {{{-4.9t^2 +20t = 5}}}
--> {{{-4.9t^2 +20t -5 = 0}}}
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Use quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}} 
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--> {{{t = (-20 +- sqrt( (-20)^2-4*(-4.9)*(-5) ))/(2*(-4.9))}}}
--> {{{t = (-20 +- sqrt(302))/(-9.8)}}}
--> {{{t = (20 +- sqrt(302))/9.8 }}}
We want the time when the stone is on its way down which will be the longer time
--> {{{t = (20 + sqrt(302))/9.8 = 3.814}}}
Trip took 3.814 sec
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v(t) = -9.8t + 20
Solve for when t = 3.814
--> {{{v = (-9.8)(3.814) + 20 = -17.378}}}
Speed is always positive
Stone was traveling at 17.378 m/s when caught