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Calculus Physics Artillery Problem (Calculus BC)

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data

    In a paper presented in 1686, Edmond Halley summarized the laws of gravity and projectile motion. These laws have since been applied to military gunnery, especially in the fields of land artillery and naval gunfire. Several problems of projectile motion were posed related to military gunnery, specifically (1) firing a projectile down an incline plane and (2) firing a projectile up an incline plane. Both of these type problems would be crucial to model for the Army, as military terrain is rarely flat. artillery projectile motion must account for variations in terrain elevation. To simplify calculations below, we will ignore air resistance, wind speed, and all other factors affecting projectile motion except gravity.

    Angle = [itex]\alpha[/itex]
    Initial Velocity= V0
    Angle of Depression = θ

    Requirement #1

    A 155mm artillery round (projectile) is fired from a self-propelled howitzer (located at the origin) down a hill slope (incline plane) that makes an angle θ with the horizontal ground. The angle of elevation of the howitzer and the initial speed of the artillery round are represented as [itex]\alpha[/itex] and V0, respectively.

    A. Find the position vector, r(t), of the artillery round (projectile). Determine the horizontal component as a function x = f (V0, [itex]\alpha[/itex], t) and the vertical component as a function y = g (V0, [itex]\alpha[/itex], t).

    B. Determine the parametric equations for the path of the artillery round as a function of time. Identify the horizontal parametric equation as a function x = f (V0, [itex]\alpha[/itex], θ) and the vertical parametric equation as a function y = g (V0, x, [itex]\alpha[/itex]), and t = f g (V0, [itex]\alpha[/itex], θ) [Solved for y first and then substituted in to find x, followed by using functions for t.]

    Requirement #2

    Artillery personnel are interested in determining the maximum range (horizontal distance) and maximum height (vertical distance) for their weapon system. Using the position vector and parametric equations derived in Requirement #1 above, determine the following:

    A. Determine the angle of elevation ([itex]\alpha[/itex]) to achieve maximum range. The angle of elevation will be a function of angle of depression of the inclined plane (θ). Describe this angle of elevation in relationship to the angle of depression.

    B. Calculate the maximum range of an artillery round with a known initial velocity of 625 meters per second and an angle of depression with the inclined plane (hill) of [itex]\pi[/itex]/8 radians.

    C. Determine the angle of elevation ([itex]\alpha[/itex]) to achieve maximum height from the horizontal and calculate the maximum height of a mortar round with a known initial velocity of 300 meters/second.

    2. Relevant equations

    g=-9.80665 meters/second
    No relevant equations due to the fact that the problem asks for equations.

    3. The attempt at a solution

    Requirement #1. A.

    r(t) = [V0cos([itex]\alpha[/itex])]ti + [(-g/2)t2 + V0sin([itex]\alpha[/itex])t]j

    x = V0cos([itex]\alpha[/itex])t

    y = (-g/2)t2 + V0sin([itex]\alpha[/itex])t

    Requirement #1. B.

    y = tan([itex]\alpha[/itex])x - (g/2)(x2/(V02cos2([itex]\alpha[/itex])))

    x = (V02sin(2α) + 2V02tan(θ)cos2([itex]\alpha[/itex]))/g

    t = (2V0(sin([itex]\alpha[/itex]) + tan(θ)cos([itex]\alpha[/itex])))/g

    Requirement #2. A.

    Range = ((2V02cos([itex]\alpha[/itex]))/(g cos2(θ))) * sin([itex]\alpha[/itex]+θ)

    dR = ((2V02)/(g cos2(θ))) [cos([itex]\alpha[/itex])cos([itex]\alpha[/itex]+θ) - sin([itex]\alpha[/itex])sin([itex]\alpha[/itex]+θ)]

    dR = ((2V02)/(g cos2(θ))) [cos(2[itex]\alpha[/itex]+θ)]

    θ = ([itex]\pi[/itex]/2)-2[itex]\alpha[/itex]

    [itex]\alpha[/itex] = ([itex]\pi[/itex]/4)-(θ/2)

    Requirement #2. B. C.

    Wanted confirmation on previous problems before moving on.
     
  2. jcsd
  3. Dec 11, 2011 #2
    Found Requirement #2. B.
    Maximum Range is at angle of 3[itex]\pi[/itex]/16 radians and is at the point (58428.86455, -22944.96145)

    Time is 112.43488 seconds after initial firing.

    Need help with the confirmation of the time function. I have solved for it twice and reached the same results. However the time does not match up with those found analytically.
     
    Last edited: Dec 11, 2011
  4. Dec 11, 2011 #3
    Requirement #2. C.
    [itex]\alpha[/itex] for maximum elevation is 90o because when θ is -[itex]\pi[/itex]/2 radians the vertical velocity is achieved because [itex]\pi[/itex]/2 is a vertical line. The Max is at t = 30.59149 seconds at a height of 4588.72296.
     
  5. Dec 11, 2011 #4
    New Revelation: on Requirement #2. B. to understand t you must solve for x and y rather than trying to figure out t initially.
     
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