Projectile Motion - rocket launched, engine, fail

  • Thread starter dandy9
  • Start date
  • #1
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Homework Statement


A rocket is launched at an angle of 58.0° above the horizontal with an initial speed of 98 m/s. It moves for 3.00 s along its initial line of motion with an acceleration of 28.0 m/s2. At this time its engines fail and the rocket proceeds to move as a free body.

Maximum altitude?
Total flight time?
Horizontal range?

Homework Equations





The Attempt at a Solution



I'm really stuck - I don't need an answer, just some help to get started... This is due in an hour so any help is appreciated.
 

Answers and Replies

  • #2
157
0
Solve it in two steps. Step one is the rocket takes off with the given velocity and acceleration provided. Step two after the rockets shut off it is now a standard projectile motion problem where it has some initial height & velocity as solved for in part 1.
 
  • #3
28
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I got 1429.7m which is wrong...
My 'step 1' height is 375,
my 'step 2' height is 1429.
Any help please?
 
  • #4
157
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How did you find your step 1 height? I got 356.18m
 
  • #5
28
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Well, I did it again and I got 355.6
I used the equation dy =viy*t+.5ay*t^2 (the "y"s are all subscripts indicating the vertical component (the "i" indicates the initial value))
which came to be dy = 83*3+.5*23.7*3^2
equaling 355.6

For the second step height I found the time it took from the moment the engine shut off to apex, which turned out to be 15.7s. I also found the final vertical velocity at that moment, which I got to be 154.2m/s.
I plugged these numbers into the formula:
dy = viy*t+.5*ay*t^2
And got 2349.9m...

but I'm thinking that this doesn't really make sense... shouldn't the rocket go higher in step 1 with the engine? I feel like my step 2 height is way too big.

Any thoughts?
Thanks,
D
 
  • #6
157
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For the second step height I found the time it took from the moment the engine shut off to apex, which turned out to be 15.7s. I also found the final vertical velocity at that moment, which I got to be 154.2m/s.
I plugged these numbers into the formula:
dy = viy*t+.5*ay*t^2
And got 2349.9m...
The vertical velocity at the moment of apex is zero. The value of 154.2m/s is the initial vertical velocity of the second step.

then you simply use:

h=v0*t+.5*a*t^2+h0 (don't forget the initial height h0)
 
  • #7
28
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AWESOME! THANKS! (The answer is 1568.7m)
I realized I had the same equation as you for step 2, just forgot to square the time - just shows how one tiny error can really mess you up.
Thanks again, you were a great help.
Much thanks,
D
 
  • #8
157
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Haha, yeah those small errors will get you... Glad I could help though.
 

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