Projectile Motion with acceleration phase

AI Thread Summary
A rocket accelerates for 2 seconds with a time-dependent acceleration of a = 2t + 4t³ at a 60º angle before coasting to impact. The initial velocity at t=2 seconds is calculated to be 20 m/s, with components of 10 m/s horizontally and 17.3 m/s vertically. The discussion revolves around finding the maximum height and impact velocity, with a focus on whether to integrate the velocity or acceleration function for height calculation. Participants clarify that gravity's effect is ignored during the acceleration phase, treating it as a rail launcher model. The consensus is to use the integral of the velocity function to determine the distance traveled.
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
\int^{2}_{0}(2t + 4t3)
[t2+t4]^{2}_{0}=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.
\int^{2}_{0}(20)
[20t]^{2}_{0}
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-
\int^{2}_{0}(t2 + t4)
[(1/3)t3 + (1/5)]^{2}_{0} = (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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