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bibby Group Title

(a) A rocket is fired straight up from a tower 110 ft above the ground with an initial velocity of 224 ft/s. (Assume the acceleration due to gravity is −32 ft/s2). When does the rocket reach its maximum height? and at this time, how far is the rocket above the ground?

  • one year ago
  • one year ago

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  1. bibby Group Title
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    @zepp @AravindG @tcarroll010

    • one year ago
  2. richyw Group Title
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    integrate the acceleration with respect to time to get the velocity, the constant of integration is the initial velocity. then integrate this velocity function with respect to time, the constant of integration will be the initial height. then you have the position as a function of time. to figure out the maximum height you just need to solve for when v(t)=0, then plug that t value into the position function to get the height!

    • one year ago
  3. bibby Group Title
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    also one more thing. When it asks for "find displacement when acceleration is 0", dpo I also plug back into s(t)?

    • one year ago
  4. richyw Group Title
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    uh, they told you to assume that acceleration is -32...

    • one year ago
  5. richyw Group Title
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    maybe that's after it hit the ground?

    • one year ago
  6. bibby Group Title
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    A different problem.

    • one year ago
  7. zepp Group Title
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    Okay, this is a physics question, but whatever. Given the initial height, let's call it \(\large y_0\), 110 ft; We are looking for the final height, \(\large y\). We also know that when the rocket will reach its highest point, it's velocity is 0, \(\large v=0\), and the acceleration would be \(\large -32ft/s^2\). From the free-fall equations of kinematics, using the formula: \(\large v^2=v_0^2+2a(y-y_0)\) So, \(\large 0=(244ft/s)^2-2a(y-110ft)\\ \large 59536=2a(y-110)\\\large 930.25=y-110\\\large y=930.25+110=1040.25\) The final height would be 1040.25 feet in the sky.

    • one year ago
  8. zepp Group Title
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    If you want to find \(t\), then plug everything in \(\large y=y_0+\frac{1}{2}(v_0+v)t\) and solve for \(t\).

    • one year ago
  9. bibby Group Title
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    zepp: that's the thing. It's one of my previous calc finals and I've never seen it worded like this although I'm pretty sure integration is how to go about it.

    • one year ago
  10. zepp Group Title
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    Of course, it would be solved using integration, but that's a classical mechanics introductory course question ;x

    • one year ago
  11. bibby Group Title
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    I guess I meant the technique and wording and not solving. I'm sure derivatives aren't exclusive to calculus :D.

    • one year ago
  12. richyw Group Title
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    \[a=\frac{dv}{dt}\]\[\int a\,dt=\int dv\]\[v=at+v_0\]\[v=\frac{dx}{dt}\]\[\int (at+v_0)dt=\int dx\]\[x=\frac{1}{2}at^2+v_0t+x_0\]

    • one year ago
  13. zepp Group Title
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    ^ could easily be derived using simple algebra :D

    • one year ago
  14. richyw Group Title
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    show me?

    • one year ago
  15. zepp Group Title
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    Let's go from the acceleration, which is \[\large a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t_f-t_i}\]If we set \(t_i=0\) and \(t_f=t\), we get \(\large a=\frac{v-v_0}{t}\), or \(\large v=v_0+at\) Now the formula to find the displacement is \(\large \Delta x=x-x_0\)

    • one year ago
  16. zepp Group Title
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    YES FINALLY IT LOADED

    • one year ago
  17. zepp Group Title
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    |dw:1355785810205:dw| We know that the displacement could be given by \(\large \Delta x = v_{av}\Delta t=\frac{(v_f - v_i)}{2}\) So the distance would be \(\large x= x_0 + \frac{1}{2}(v_0+v)t\) Now, if you substitute \(v=v_0+at\) into the formula above, you'll get \(\large x = x_0+v_0t+\frac{1}{2}at^2\) by algebraic manipulations :)

    • one year ago
  18. bibby Group Title
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    ah sorry I closed the question. I don't think pm will work out but good luck if you guys can still type

    • one year ago
  19. richyw Group Title
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    you can type after it's closed anyways! and yeah but this works because acceleration is constant right? personally i'd just do it the faster way! (i can never memorize the kinematic eqns though)

    • one year ago
  20. zepp Group Title
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    Yes, acceleration is constant, it's the gravitational pull, usually, so :P

    • one year ago
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