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Cannon fired from moving truck

  1. Oct 11, 2008 #1
    Hello everyone,

    I have a physics question for you all. A cannon fired from a truck at 5m/s. A cannon is fired in the same direction as the cannon. The velocity of the projectile is 50m/s. What angle will provide the greatest (distance in the x-driection) range for the projectile.

    I have gotten two solution.
    Using triangles and that both have the same maximum height the best angle is 49 degrees.

    Using calculus we have come with a best angle of 46.9 degrees.

    Which is best and why? I believe it is calculus because the calculus version is assuming the cannon fired at the non-45 degrees will have a higher altitude. Therefore ruining the triangles method.

    Thank you.
  2. jcsd
  3. Oct 11, 2008 #2

    Doc Al

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    I have no idea what you mean by using "triangles". Clearly at least one of these methods must be wrong if they get different answers, so they must be based on different assumptions. Describe both solutions.
  4. Oct 11, 2008 #3
    The triangles method uses a 45-45-90. From the top angle drop a line down to the base leg. The left side is 5 and the other is x. The side in common is h, the maximum height. Since this is a 45-45-90, the 5+x side is equal to the h. There is enough variables equal to each other that the angle in the x-h triangle can be found to be 49 degrees.

    However, change it to using the x=vt equation and things change. Since we are giving the projectile the extra x-velocity I wrote the velocity in the x-direction to be:

    vx= vtruck+vcos(theta) v=the velocity of the projectile


    I want to do a derivative on theta and find the zero to get the max distance. So I need to get rid of t.

    t is found using vf=vi+at. All of this in the y-direction

    t=2vsin(theta)/g again v is projectile velocity

    So after some substitutions, cleaning up, taking a derivative and getting a sin^2 and a cos^2, I found an equation with thetas hard to pull out.

    I put the equation into excel and forced angles into the equation until my derivative equation hit zero, therefore, showing me the angle that creates the maximum distance.

    This method gives 46.9 as theta.

    I hope this clears up what I was asking. Thank you for your time.

  5. Oct 12, 2008 #4

    Doc Al

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    Since the trajectory of the cannonball is not a triangle, this method doesn't make any sense to me.

    Looks good to me. Just for the challenge of it, try to solve for theta analytically. (Express that last equation completely in terms of cosines. You'll get a quadratic that you should be able to solve easily. You'll get the same answer of course.)
  6. Oct 12, 2008 #5
    Good morning,

    Thanks for looking in to it. I will see what I can do with the cos's.

  7. Oct 12, 2008 #6
    First of all, maximum horizontal projectile range is always 45 degrees.

    Second, the horizontal range is given by (v^2/g) Sin 2(theta)...

    these can be derived from x (range) =vt cos(theta)

    and y (vertical) = vtsin(theta) - 1/2 gt^2

    and of course "t" time of flight is the same for both x and y...
  8. Oct 12, 2008 #7
    Max range is always at 45 when you projectile is launched from an object that is not moving in the x-direction.

    If you fire a cannon that is on a railcar moving with a velocity, a 45 degree cannon will not give a max range. Since there is already a x-velocity you can then steal some x-velocity from the cannon and give it to the y-velocity; therefore, giving it more time in air; therefore, more time to run in the x-direction.
  9. Oct 12, 2008 #8

    Doc Al

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    Only for a fixed speed (v), which is not the case here. (The speed of the cannonball with respect to the ground depends upon the angle.)
  10. Oct 12, 2008 #9


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    Can we assume that you mean the TRUCK is moving at 5 m/s and that the cannon is fired in the same direction as the truck is moving?

    I don't get "49" or "46.9" degrees. I get approximately 45.17 degrees.
    Last edited by a moderator: Oct 12, 2008
  11. Oct 12, 2008 #10

    Doc Al

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    Yes. And that the cannonball speed is 50 m/s with respect to the truck.
    Do it over. :wink:
  12. Oct 12, 2008 #11
    Hello HallsofIvy,

    The cannon is on the truck. The truck is moving at 5m/s, lets say moving to the east. The cannon is fired with an unknown angle. The projectile has a velocity of 50 m/s at the unknown angle. The cannon is fired in the east direction.
  13. Oct 12, 2008 #12
    just add 5 ms/sec cos(theta) and 5m/s sine (theta) to the fifty m/s to determine total projectile speed....for altitude or range or whatever you want to play with.....the projectile doesn't know if the cannon is moving or if it has a slightly bigger charge...
  14. Oct 12, 2008 #13
    The velocity vector for the projectile on a moving railcar, truck, whatever is going to be

    v = [5+vcos(theta)]i + [vsin(theta)]j

    The 5m/s is added to the i component (x-dir). You can't go and add 5m/s to the y-dir.
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