Question 314607
When ball hits the ground, h(t)=0. Hence,

h(t)=-16t^2+64t+5=0

or 16t^2 -64t -5=0 , a simple quadratic eqn.

Putting the appropriate values in the standard solution of a Quadratic Equation

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}  ,  we get ,


t = [ - ( - 64)  + - sqrt { ( - 64)^2 - 4*16*( - 5)} ] / (2 * 16)

t = [ 64  + - sqrt { 64 ( 64 +  5} ] / (2 * 16)

t = [ 64  + - 8 * sqrt ( 69) ] / (2 * 16)

t = [ 8  + - sqrt ( 69) ] / 4

As t can not be negative and sqrt(69) > 8 , so the only acceptable solution for t is 

t = [ 8  + sqrt ( 69) ] / 4

 t=16.31 sec.