Homework Help: Projectile Motion with acceleration phase

1. Sep 18, 2008

ionic_scream

1. The problem statement, all variables and given/known data
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

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

3. 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

2. Sep 18, 2008

LowlyPion

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?

3. Sep 18, 2008

ionic_scream

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.

4. Sep 18, 2008

LowlyPion

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.