Question

A rocket is launched at t=0 seconds. It's height, in feet, above sea-level, as a function of time,
t, is given by

h(t)=-16t2+64t+192

When does the rocket hit the ground after it is launched?

Answer 1

-16(t^2-4t-12) =0
(t-6)(t+2) =0
t - 6 = 0
t = 6

Answer 2

V = Vo + g*Tr.
0 = 64 - 32Tr, Tr = 2 s. = Rise time.

h = -16Tr^2 + 64Tr + 192.
h = -16*2^2 + 64*2 + 192 = 256 Ft.

h = 0.5g*Tf^2.
256 = 16Tf^2, Tf = 4 s. = Fall time.

Tr+Tf = 2 + 4 = 6 s. To hit gnd.