(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A rocket is launched and accelerates for 2 seconds with an acceleration a = 2t + 4t^{3}at 60º wrt the horizontal, it then "coasts" until impact at the same altitude which it was launched from.

Find a)Vat t=2s

b) Max Height

c) Impact Velocity

2. Relevant equations

Y - Y_{0}= V_{y0}t + .5at^{2}

V_{y}- V_{y0}= 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 + 4t^{3}

[tex]\int^{2}_{0}[/tex](2t + 4t^{3})

[t^{2}+t^{4}][tex]^{2}_{0}[/tex]=20

So V(2) = 20 m/s

Sin(60º) = y_{1}/20

y_{1}=17.3 m/s

Cos(60º) =x_{1}/20

x_{1}= 10 m/2

SoV= (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](t^{2}+ t_{4})

[(1/3)t^{3}+ (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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# Homework Help: Projectile Motion with acceleration phase

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