After how many seconds does the land on the ground?

Question

You throw a ball up into the air with an initial vertical velocity of 46 feet per second. The ball leaves your hand when it is 6 feet above the ground.

Answer 1

h(t)=-16t^2+vt+h where v=velocity in feet per second and h=height above the ground from where the ball is thrown up and h(t)=height in feet above the ground which is zero

0=-16t^2+46t+6

16t^2-46t-6=0

divide all terms by 2

8t^2-23t-3=0

factor

(8t+1)(t-3)=0

t-3=0

t=3 seconds  (8t+1=0 gives us a negative answer)