A coin is tossed upward from a balcony 400 feet high with an initial velocity of 16 feet per second. During what interval of time will the coin be at a height of at least 80 feet?

Answer 1

80 = -16t^2 + 16t + 400

divide by 8

10= -2t^2 + 2t + 50

2t^2 -2t = 40

divide by 2

complete the square, add 1/4 to both sides

t^2 -t + 1/4 = 40/2 + 1/4

(t- 1/2)^2 = 81/4

t-1/2 = +9/2 (ignore the negative square root)

t = 1/2 +9/2 = about 5 seconds is when it's at 80 feet

and 5 seconds is the time when it's above 80 feet

or you could have factored t^2-t-20

(t-5)(t+4) = 0

t-5 =0

t = 5 seconds

it's theoretically at 80 feet also when t=-4

then you'd have it above 80 feet for 5--4=5+4=9 seconds

but time starts at t=0, usually no negative time allowed