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bibby

  • 2 years ago

(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?

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  1. bibby
    • 2 years ago
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    @zepp @AravindG @tcarroll010

  2. richyw
    • 2 years ago
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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!

  3. bibby
    • 2 years ago
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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)?

  4. richyw
    • 2 years ago
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    uh, they told you to assume that acceleration is -32...

  5. richyw
    • 2 years ago
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    maybe that's after it hit the ground?

  6. bibby
    • 2 years ago
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    A different problem.

  7. zepp
    • 2 years ago
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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.

  8. zepp
    • 2 years ago
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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 \(t\).

  9. bibby
    • 2 years ago
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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.

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

  11. bibby
    • 2 years ago
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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.

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

  13. zepp
    • 2 years ago
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    ^ could easily be derived using simple algebra :D

  14. richyw
    • 2 years ago
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    show me?

  15. zepp
    • 2 years ago
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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 \(\large \Delta x=x-x_0\)

  16. zepp
    • 2 years ago
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    YES FINALLY IT LOADED

  17. zepp
    • 2 years ago
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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 :)

  18. bibby
    • 2 years ago
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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

  19. richyw
    • 2 years ago
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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)

  20. zepp
    • 2 years ago
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    Yes, acceleration is constant, it's the gravitational pull, usually, so :P

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