A test rocket is fired straight up from rest. The net acceleration is 20 m/s2 upward and continues for 4.0 seconds, at which time the rocket engines cease firing. What maximum elevation does the rocket reach?

Answer 1

well, do the power thing first
v = 0 + a t
v = Vi = 20 * 4 = 80 m/s at power cutoff
at what height did the power stop?
h = 0 + 0 t + (1/2)(20)(16)
= 160 meters
so
we have Vi = 80 m/s at Hi = 160 meters
now coast up
v = Vi - g t
at top v = 0 , using g = 9.8
0 = 80 - 9.8 t
t = 8.16 seconds coasting to peak
then
h = Hi + Vi t - 4.9 t^2
h = 160 + 80 (8.16 = 4.9(8.16)^2