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Projectile Motion Help!

  1. Dec 14, 2004 #1
    This is a very important question, I need details on how to solve it step by step.

    A cannon on the Bismark German Battleship, fires a 2000 lbs (907kg) shell at an angle of 40degrees. It travels a distance of 20 miles (32.2km), what is the speed of the shell leaving the cannon? What is the variable t?

    Like I said this is a very important problem that I need to solve for my Honors Senior Physics class.

    Lemme show what I think I know.

    Theta = 40 degrees
    m = 2000lbs (907kg)
    d = 20mi (32.2km)
    a = -9.8 m/s^2 (acceleration due to gravity?)
    vi = Initial velocity is equal to 0
    t = ???
    W = weight is equal to 9.8 x 907000g ?
    netF = -9.8 x 907 kg?

    Please correct me where I am wrong, and show me what is needed and how I figure this problem out. I was told I need no outside information such as information detailing the German battleship.
  2. jcsd
  3. Dec 14, 2004 #2
    You have all the information you know. First off, you know that it travels 20 miles. As a result, you know that it spends 10 miles traveling up, and 10 miles traveling downwards. Remember, horizontal velocity stays constant. Now see if you can find an equation to help you out to see how much time it takes you.

    The other thing you can do is to try to derive the trajectory/range equations and play around with that. That would save you all the intermediate steps....
  4. Dec 15, 2004 #3
    Still completely and utterly lost, the horizontal velocity is a constant. 0-10 miles are spent going up while everything after 10 miles is spent goin down. Can anyone help me out some more?
  5. Dec 15, 2004 #4
    F=-mg *e _y

    [tex]v_x = v_{0x}[/tex]
    [tex]x-x_0 = v_{0x} * t[/tex]
    [tex]v_y = v_{0y} -gt[/tex]
    [tex]y-y_0 = v_{0y}*t -gt^2/2[/tex]

    v_ox and v_oy are the initial horizontal and vertical velocity...These are the equations that you will be needing...For example you know that the object travelled a given distance...try to put this into these equations....

  6. Dec 15, 2004 #5


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    Initial velocity is not equal to 0.

    As others said, you need to break this up into two equations: one for horizontal and one for vertical. While you can't solve either one directly, you can establish a relationship between vi_x and t. If you substitute that relationship into your equation for vertical motion only once, you'll see the problem is very easy to solve. Even if you get carried away and substitute all you can, the problem's still solvable.

    Hint: [tex]tan \theta = \frac{y}{x}[/tex]
  7. Dec 15, 2004 #6


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    In addition to above, the initial velocity would be zero iff you were working with the system inside the cannon barrel. Then the final velocity would be the velocity of the ball when leaving the cannon. Which you could in turn use as your initial velocity for the projectile motion. You can only do this though, if you know the force exerted by the impulse of the blast.

    Else the above equations and explanation are sufficient advice to solve the equation.
  8. Dec 15, 2004 #7
    Alright as I said I am totally lost right now, my physics teacher uses different variables so it seems different for me. Is it possible for someone to solve the equation for me and walk me step by step through it so I can understand it better?
  9. Dec 16, 2004 #8


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    I don't know what's the m = 2000lbs (907kg) is used for. But here's my suggestion:
    [tex]x_{max} = \frac{v_{initial}^2 \sin{2 \theta}}{g}[/tex]
    Try to prove this yourself.
    [tex]x_{max} = 20 mi [/tex]
    You will work out
    From there you will work out the variable t.
    Hope this help and... hope I am right, :-)
    Viet Dao,
  10. Dec 16, 2004 #9


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    VietDao29's method certainly is correct. However, I think the equation most appropriate for what you want to do is:

    [tex]s_f=s_i + v_i t + \frac{1}{2} a t^2[/tex]

    [tex]s_f[/tex] is final position
    [tex]s_i[/tex] is initial position
    [tex]v_i[/tex] is initial velocity (when leaving cannon)
    [tex]a[/tex] is acceleration (gravity acclerates an object downward, negative)

    You need to use this equation for both the vertical component and the horizontal component. There is no acceleration in the horizontal direction.
  11. Dec 16, 2004 #10
    I'm also pretty sure VietDao29 is right. THat equation is derived from the relations between the x component veloctiy vector and the y-component at 40degrees, right? With it, you should be able to find x component and use everyones favorite physics formula [tex] d = v_c t[/tex] to find t, right?
  12. Dec 16, 2004 #11
    Can anyone answer vietdao29's formula? Can you tell me what the formula is and actually solve it? I dont believe I am solving it right because for an initial velocity leaving the cannon I got 199 (not sure if its miles per hour or what).. Can anyone tell me exactly what I need to solve my question, what is the speed leaving the cannon?
  13. Dec 17, 2004 #12


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    Try again proving it. It is not very hard.
    If you still cannot work it out, just post here and I will help you.
    You can use metters per second, or miles per second in this formular. If you use g = 9.81 m / s^2, then it must be metters per second.
    Bye bye,
  14. Dec 17, 2004 #13


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    I'm stumped how you could get 199.

    You have to rearrange VietDao29's formula to isolate v

    [tex]x_{max} = \frac{v_{initial}^2 \sin{2 \theta}}{g}[/tex]


    [tex]\sqrt{\frac{x_{max} g}{sin {2 \theta}}}=v_i[/tex]

    Once you get your total velocity, you have to break it into its components to solve for time.
  15. Dec 17, 2004 #14
    I ended up with 14.1 meter per second I believe? I took distance times gravity, then took sin (2 x 40) (yes I put it in like that not sure if im right) then took the answer from D x G (196) and divided it by .985, ended up with 198.9 then square rooted that. Resulting in 14.1, but my physics teacher had told me it was not correct, is there something I am not getting? This is a very vital problem I have to solve, my physics teacher has gotten very ill and is to stop teaching and I feel I owe it to him to solve it (nobody else in the class has even come close). Help me out! Also, is that all I need to do to find the speed of the shell leaving the cannon or is there more to do, that could be why he told me it was wrong.
  16. Dec 17, 2004 #15


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    There's your problem. You multiplied 20 miles by 9.8 meters/sec^2. You need to use the same units all the way through.
  17. Dec 17, 2004 #16
    So should I convert the 20 miles to kilometers or meters and use it in the problem?
  18. Dec 18, 2004 #17


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    You must convert it to metters if you use g = 9.8 m / s ^ 2. And therefore, it will return you the speed in m / s.
    You cannot use km with the g in m/s ^ 2, or miles with the g in m / s ^ 2. It makes no sense. It must be m with g in m / s ^ 2.
    So 20 mi = 32,2 km = 32200 m
    Is it all clear, now?
  19. Dec 18, 2004 #18
    566 m/s seem about right? Thats the speed leaving the cannon if I am correct. Which would also be 1267.2 mi/h ??? Does it seem correct?
  20. Dec 18, 2004 #19


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    Yes, that is your velocity as the shell leaves the cannon. Getting the first of the unknowns is just the hard part. The question also asked how much time it took for the shell to complete its trajectory.

    You have to break the velocity vector up to get that. Horizontal velocity being v cos theta and vertical being v sin theta.
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