A model rocket is fired vertically and ascends with a constant vertical acceleration of 4.00m/s^2 for 6s...?

Question

Its fuel is then exhausted and it continues as a free-fall particle. (A) What is the max altitude reached? (B) What is the total time elapsed from takeoff until the rocket strikes the Earth?

Answer 1

D = 1/2 a (t^2) = (1/2)(4)(36) = 72 m (assuming it started off from rest)

Second part is the same equation

D = 1/2 a (t^2)  where a= 9.8 m/s^2

t = [2D/a]^(1/2)

t= 3.83325939 s

Answer 2

At point where fuel is exhausted

s=1/2*at^2=1/2*4*(6)^2=72 meter

and it's velocity at this time is: Vi=a*t=24m/s

It continues to rise for another:

0=Vi-gt

t=-Vi/-g=(-24/-9.8)

=(-24) / (-9.8) = 2.45 s

it's average velocity during this rise time is

(24-0)/2=12 m/s

and it travels the extra distance:

s2=Vave*t=12 * 2.45 = 29.4 meter

max altitude= s1+s2=72 + 29.4 = 101.4 meter

time up=6+2.45=8.45 seconds

from s=1/2*g*t^2

time down=sqr(2 * 101.4 * 9.8) = 44.58 seconds

total time=44.58 + 8.45 = 53.03 seconds