Question 255611
The initial height a ball is thrown in the air is 80 feet, velocity 64 ft. per sec. and g is 32 feet/sec squared. I am supposed to figure out when it will hit the ground in 2 different ways. 
---------------
h(t) = -16t^2 + vot + so
---
The height is zero when the ball hits the ground:
Solve -16t^2+64t+80 = 0
Divide thru be -16 to get:
t^2 - 4t -5 = 0
Factor:
(t-5)(t+1) = 0
Positive solution:
t = 5 seconds
-------------------------
Graph the equations to see when the ball hits the ground:
{{{graph(400,300,-10,10,-10,140,-16x^2+64x+80)}}}
===================================================
Cheers,
Stan H.