Projectile Motion with acceleration phase

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Homework Help Overview

The problem involves a rocket launched with a time-dependent acceleration at an angle, followed by a projectile motion phase. Participants are tasked with finding the velocity at a specific time, the maximum height, and the impact velocity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the method of finding the velocity vector at the end of the acceleration phase and question the correct approach to calculate the height. There is uncertainty about whether to integrate the velocity directly or to consider the original acceleration function.

Discussion Status

Some participants have provided guidance on treating the problem as a rail launcher and suggested that the effect of gravity is ignored during the acceleration phase. There is acknowledgment of the need to clarify how the angle of launch affects the calculations.

Contextual Notes

Participants note that the problem constraints may lead to confusion regarding the treatment of gravity and the angle of launch during different phases of motion.

ionic_scream
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Homework Statement


A rocket is launched and accelerates for 2 seconds with an acceleration a = 2t + 4t3 at 60º wrt the horizontal, it then "coasts" until impact at the same altitude which it was launched from.
Find a) V at t=2s
b) Max Height
c) Impact Velocity


Homework Equations


Y - Y0 = Vy0t + .5at2
Vy - Vy0 = at



The Attempt at a Solution


(a)
First I need to find the velocity in vector notation for t = 2, which will also be the initial velocity for the projectile motion section of the problem.
a = 2t + 4t3
[tex]\int^{2}_{0}[/tex](2t + 4t3)
[t2+t4][tex]^{2}_{0}[/tex]=20
So V(2) = 20 m/s

Sin(60º) = y1/20
y1=17.3 m/s

Cos(60º) =x1/20
x1 = 10 m/2

So V = (10m/s)i + (17.3m/s)j
(b)
To start the projectile motion problem I first need to find the initial height(The height at which the acceleration phase ends)
So I took the integral of the velocity at 2s to find the position at 2s.
[tex]\int^{2}_{0}[/tex](20)
[20t][tex]^{2}_{0}[/tex]
So r(2) = 40

This is where my question is, is this correct to take the integral of the result of part a velocity? Or should I take the integral like the following?

-OR-
[tex]\int^{2}_{0}[/tex](t2 + t4)
[(1/3)t3 + (1/5)][tex]^{2}_{0}[/tex] = (8/3) + (32/5)

? I don't have any problems working out the rest of the problem I just can't remember the proper procedure to find the height? Any help would be awesome since I'll be tested on this in less than 24 hours! haha thanks
 
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ionic_scream said:

Homework Statement


A rocket is launched and accelerates for 2 seconds with an acceleration a = 2t + 4t3 at 60º wrt the horizontal, it then "coasts" until impact at the same altitude which it was launched from.
Find a) V at t=2s
b) Max Height
c) Impact VelocityI don't have any problems working out the rest of the problem I just can't remember the proper procedure to find the height? Any help would be awesome since I'll be tested on this in less than 24 hours! haha thanks

My concern with your approach is that I'm not sure that you have accounted for the 60 degree angle of launch.
Is the effect of gravity on the y component to be ignored?
Is the initial velocity taken at the end of 2 seconds for the free fall gravity phase still at the angle of 60 degrees?
 
LowlyPion said:
My concern with your approach is that I'm not sure that you have accounted for the 60 degree angle of launch.
Is the effect of gravity on the y component to be ignored?
Is the initial velocity taken at the end of 2 seconds for the free fall gravity phase still at the angle of 60 degrees?

Yea we are just supposed to ignore gravity's effect during the acceleration phase and assume it doesn't start until free fall phase. It sounds dumb i know, but maybe when my instructor wrote it he thought the given acceleration has accounted for it's effect? idk, but yea it's still at 60º at the start of free fall.
 
ionic_scream said:
Yea we are just supposed to ignore gravity's effect during the acceleration phase and assume it doesn't start until free fall phase. It sounds dumb i know, but maybe when my instructor wrote it he thought the given acceleration has accounted for it's effect? idk, but yea it's still at 60º at the start of free fall.

So then treat it as a rail launcher I guess is the model to follow. Then the rest is simple kinematics as you note. Otherwise I don't see anything that stands out in error in your technique.

I would choose to take the integral over the Velocity function itself to get the distance traveled.

Good Luck.
 

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