Rocket Kinematics: Acceleration and Maximum Altitude

AI Thread Summary
The discussion focuses on determining the acceleration of a rocket during its burn phase to ensure equipment survival and calculating its maximum altitude. The rocket accelerates for 30 seconds before entering free fall for the remaining 270 seconds of its total 300-second flight. The equations of motion are applied to both phases, with the first phase using the formula x = x0 + v0t + (1/2)at^2 to find the altitude reached during the burn. The second phase accounts for free fall, where the initial position is the altitude from phase one and the final position is zero at ground level. The calculated acceleration for the rocket during the burn phase is 41.8 m/s², which is confirmed as valid.
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Homework Statement


You must determine the acceleration of a rocket so that its equipment can be designed to survive. The rocket will have a burn time of t = 30 seconds, during which time it flies has a constant acceleration a. Call this Phase 1. After the fuel is exhausted the rocket enters free fall. Call this Phase 2. The total flight time is 300s.

a) what should you make the acceleration of the rocket a when the engine is on?
b)what is the maximum altitude of the rocket

//So I don't even know where to start or what part a is asking. What is the condition for the equipment to survive?


Homework Equations


x=x0+v0t+(1/2)at^2


The Attempt at a Solution



I'm lost...
 
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I think none of the equipment survives at the end of the free-fall phase, when t= 300s--so don't worry about that.
 
Yeah that is what I figured as well, but this doesn't really help me solve the question haha.
 
So you have two phases where the acceleration is constant, but different. You need to come up with 2 distinct [edit: sets of] equations, each describing the 2 time segments. A third equation might relate the 2 time segments to the total flight time.
 
Last edited:
Hmm okay so...
using x = x0 + v0t + 1/2at^2 for phase one i find that x = 1/2a(30)^2

Phase two I know it hits ground so final x is 0. The inital x for this phase is x from phase 1 so...

-x = V0(270s) - 1/2(g)(270)^2

V0 is the final velocity for phase 1.

V = V0 + at => phase 2 V0 = a30.

Inputing for V0 and -X in phase two I got:
-.5(a)(30^2) = a(30)(270) - (.5)(9.8)(270^2)

Solving for a I got... 41.8 m/s^2 for part a.

Does this look valid?
 
Yes, actually!
 
Thank you very much for your help! :)
 
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