rocket-max height question?

Question

A model rocket is launched straight upward with an initial speed of 70.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s2 until its engines stop at an altitude of 150 m. (a) What is the maximum height reached by the rocket?

b) How long after lift-off does the rocket reach its maximum height?
s
(c) How long is the rocket in the air?
s

Answer 1

first, we have to find the speed of the rocket at 150m.

Vf^2 = 2ad + Vi^2
Vf^2 = 2(2)(150) + (70)^2
Vf^2 = 5500
Vf = 74.16 m/s

The speed of the rocket at 150m is 74.16m/s. From there, it accelerates at -9.8m/s^2 because the engine stops. Find the distance it covers while accelerating at -9.8m/s^2

Vf^2 = 2ad = Vi^2

when it reaches its maximum point, its speed is 0m/s
0^2 = 2(-9.8)d + (71.16)^2
-(74.16)^2 = -19.6d
d = 280.61m

so the maximum height is 280.61 + 150 = 430.61m


b) Vf = at + Vi
74.16 = 2t + 70
4.16 = 2t
t = 2.08

it takes 2.08s to reach 150m

Vf = at + Vi
0 = -9.8t + 74.16
-74.16 = -9.8t
t = 7.57s

so it takes 7.57 + 2.08 = 9.65s to reach the maximunt height

c)The rocket falls from its maximun height
t = sqrt(2d / g)
t = sqrt(2*430.61 / 9.8)
t = 9.37s

so the total flight time is 9.37 + 9.65 = 19.02s

Answer 2

All answers to 3 sig figs. Additional precision is not only unnecessary, it is incorrect.

X = Vi*t + 1/2at²
150 = 70t + 1/2*2*t²
t² + 70t - 150 = 0
(t - 2.08)(t + 72.1) = 0
t = 2.08 sec (you can't have a negative time)

V(150) = Vi + at
...= 70.0 +2*2.08
...= 74.2 m/sec

Vf = V(150) - gt
0 = 74.2 - 9.81t
t = 74.2/9.81
...= 7.56 sec

Xf = X(150) + V(150)t - 1/2gt²
... = 150 + 74.2*7.56 - 9.81*(7.56)²/2
... = 431 m