Rocket landing?

Question

A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t) = 60t at which time the fuel is exhausted and it becomes a freely "falling" body. Fourteen seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to -18 ft/s in 5 s. The rocket then "floats" to the ground at that rate.

(a) At what time does the rocket reach its maximum height? (Give your answer correct to one decimal place.)
11.4 seconds.

(b) What is that height? (Give your answer correct to the nearest whole number.)
1409 ft.

(c) At what time does the rocket land? (Give your answer correct to one decimal place.)
I cannot find this for my life, if someone could show me the steps and the answer, it would be really appreciated. Oh, and the other two answers a and b have been verified correct.

Answer 1

To clarify things, let's break the trip into 4 distinct phases:

Phase 1: Rocket engine is burning (3 seconds)
Phase 2: Engine has burnt out; rocket is in free fall (14 seconds)
Phase 3: Parachute has opened and rocket is decelerating to -18 ft/s (5 seconds)
Phase 4: Rocket is floating to the ground at -18 ft/s (?? seconds)

> acceleration for the first three seconds is a(t) = 60t.

You didn't specify any units, but elsewhere you use "feet" for length. So I'm assuming that with units properly filled in, the acceleration is:

a(t) = (60 ft/s^3) (t) [for Phase 1]

You need to take the integral of that to find the velocity as a function of time:

v(t) = integral(a(t))
= (30 ft/s^3)(t²) + v0

But v0 is zero, so:

v(t) = (30 ft/s^3)(t²) [for Phase 1]

Integrate once again to get height as a function of time:

h(t) = integral(v(t))
= (10 ft/s^3)(t^3) + h0

But h0 is zero, so

h(t) = (10 ft/s^3)(t^3) [for Phase 1]

At t=3 sec, the engine burns out. At that time, the rocket is traveling at this speed:

v(3) = (30 ft/s^3)(3s)² = 270 ft/s

and its height is:

h(3) = (10 ft/s^3)(3s)^3 = 270 ft

After t = 3s, it accelerates under gravity. The time Δt it takes to decelerate from 270 ft/s to zero (maximum height) is:

Δt = (270 ft/s) / g
= (270 ft/s) / (32.2 ft/s²) = 8.4 s

So, added to the initial 3 seconds, the total time from ground to max height is:

3 s + Δt
= 3 s + 8.4 s = 11.4 s (just as you said).

The max height is:

h_max = 270 ft. + (270 ft/s)(Δt) - g(Δt)²/2
= 270 ft. + (270 ft/s)(8.4 s) − (32.2 ft/s²)(8.4 s)² / 2
= 1401.9 feet (more or less as you said).

Now for part "c".

Phase 2 lasts for 14 seconds, during which the rocket is in free fall, so we can use freefall equations for this phase.

At beginning of Phase 2, height h0 is 270 ft, and speed v0 is 270 ft/s, as calculated above:

Speed at the end of Phase 2 is:

v_end_phase2 = v0 − g(14 s)
= 270 ft/s − (32.2 ft/s²)(14 s)
= −180.8 ft/s

Height at the end of Phase 2 is:

h_end_phase2 = h0 + v0(14 s) − g(14 s)² / 2
= 270 ft + (270 ft/s)(14 s) − (32.2 ft/s²)(14 s)² / 2
= 894.4 ft

Now for Phase 3, the deceleration under parachute.

We are told that the velocity slows linearly; that means the acceleration (let's call it "a_phase3") is constant. We can calculate it as the change in velocity divided by time:

a_phase3 = (-18 ft/s − v_end_phase2) / (5 sec)
= (-18 ft/s − (-180.8 ft/s)) / (5 sec)
= 32.56 ft/s²

from that, we can calculate the distance traveled during Phase 3:

d_phase3 = (v0)(t) + (a_phase3)(t)² / 2
= (-180.8 ft/s)(5 s) + (32.56 ft/s²)(5 s)² / 2
= -493 ft.

So, its height at the end of Phase 3 is:

h_end_phase3 = h_end_phase2 + d_phase3
= 894.4 ft + (-493 ft)
= 401.4 ft.

Finally, during Phase 4, it floats the remaining 401.4 feet at a constant speed of 18 ft/s. So, the time it takes for this is:

t_phase4 = (401.4 ft) / (18 ft/s) = 22.3 seconds.

So, the total time is:

t_phase1 + t_phase2 + t_phase3 + t_phase4
= 3s + 14s + 5s + 22.3s
= 44.3 seconds

You'll have to double-check my math; but I think the basic method is correct.

Answer 2

Assuming there's a return and forth to be taken up, the rocket itself is in 3 separate areas, which detach one via one using fact the rocket passes throughout the time of the ambience into area. The areas of the rocket are dropped and fall into the sea the place they are then retrieved. The return and forth lands on the earth interior the comparable way an plane would.