- #1
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