anonymous
  • anonymous
A rocket blast off with an acceleration of 5m/s^2. Its booster last for 35s. if then shuts off How high did the rocket reach?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
how did you get that
Shane_B
  • Shane_B
Well...not quite. Assuming the net acceleration of 5m/s^2: Use the kinematic equation: \[d=v_it+\frac{1}{2}at^2\]Since the initial velocity is 0m/s you can simplify that to:\[d=\frac{1}{2}at^2\]Now plug in the values and see what you get...
anonymous
  • anonymous
would it be 30.625m

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anonymous
  • anonymous
jk
Shane_B
  • Shane_B
Check your math :)
raffle_snaffle
  • raffle_snaffle
If you are dealing with projections being shot upward or across you always need to take gravity into consideration. Well actually there is no acc. in the x direction only the y direction.
anonymous
  • anonymous
3062.5m
Shane_B
  • Shane_B
\[d=\frac{1}{2}(5m/s^2)(35s)^2=3062.5m\]If you factored in gravity...your rocket wouldn't be taking off at all...so the problem is assuming that you're actually moving upward at 5m/s^2.
anonymous
  • anonymous
what if you weren't factoring in gravity then what would the answer be
Shane_B
  • Shane_B
The one I posted :) I assumed a NET upward acceleration of 5m/s^2
Shane_B
  • Shane_B
3062.5m
anonymous
  • anonymous
so if you factored in gravity would the answer change
Shane_B
  • Shane_B
If you were to count gravity, your equation would look like this:\[d=\frac{1}{2}(5m/s^2-9.8m/s^2)(35s^2)\]That will give you a negative answer which means the rocket won't be taking off.
Shane_B
  • Shane_B
That's why I said that the the problem must be assuming that it's a NET upward acceleration of 5m/s^2....because otherwise the answer would not make sense.
raffle_snaffle
  • raffle_snaffle
changing the sign of gravity wouldn't make the answer logical?
raffle_snaffle
  • raffle_snaffle
If you took gravity to be + and upward to be - you would still obtain an answer with a - number?
anonymous
  • anonymous
oh ok
Shane_B
  • Shane_B
No...but if you did that, then you're assuming upward to be a negative direction which would result in a valid answer. Think about that for a moment :)
anonymous
  • anonymous
ok thank you
Shane_B
  • Shane_B
Maybe my last post isn't very clear. What I mean to say is that if you reverse both signs and end up with a negative answer, it would be valid since negative would be an upward movement. In that case, if your answer was positive, it would be invalid.
Shane_B
  • Shane_B
@annej> you're welcome :)
raffle_snaffle
  • raffle_snaffle
I swear, when I was using the kinematic equations in phy I would have to always use a positive g. Because the g in the formula was already turned into a + g from a - g. Lol
Shane_B
  • Shane_B
The sign of g just depends on how you set the problem up. It doesn't really matter as long as long as you stick with the signs for the directions throughout the problem. I can walk through this one real quick and flip the signs...and you'll see that it won't matter.
Shane_B
  • Shane_B
Lose your sign...lose your mind, I always say :)
raffle_snaffle
  • raffle_snaffle
Haha okay I believe you.=)

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