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A shot fired from gun, muzzle velocity of 1200 ft/s

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  1. Feb 14, 2016 #1
    • Mentor note: moved to homework section
    A shot fired from a gun with a muzzle velocity of 1200 ft per second is to hit a target 3000 feet away. Determine the minimum angle of elevation of the gun?

    Ok so I know that r(t)= (vcos(theta) t)i +(vsin(theta) t +16t^2)j

    I tried to set vcos(theta)t =3000, solve for t, and plug into the j component. However, I am getting confused with the math because i get t= 5/2cos(theta) and when I plug it in it gets pretty messed up. Can someone help?
     
  2. jcsd
  3. Feb 14, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    It should not get too messy. Please show your calculations, otherwise it is impossible to tell what went wrong.
    You either have a strange sign or angle convention with the j component.
     
  4. Feb 14, 2016 #3
    1200cos(theta) t=3000
    t=3000/1200cos(theta) or 5/2cos(theta)

    so I plug into the j component
    1200sin(theta)(3000/1200cos(theta))- 16(5/2cos(theta))^2

    That's where I'm stuck. I'm really not sure where to go from here
     
  5. Feb 14, 2016 #4

    mfb

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    2016 Award

    Staff: Mentor

    There are brackets missing for denominators, and the last expression should be equal to something. That allows to simplify it.
     
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