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Homework Help: Max horizontal displacement of a projectile

  1. Dec 20, 2012 #1
    1. The problem statement, all variables and given/known data
    A catapult is capable of firing a projectile with a velocity of 25 m/s, the catapult is surrounded by a wall of height 20.4 m what is the maximum distance measured from the wall, that the catapult
    projectiles are capable of hitting?

    2. Relevant equations
    [tex] h=(v_o^2sin^2θ)/2g[/tex]
    [tex] x=v_otcosθ [/tex]
    [tex] y=v_otsinθ -0.5gt^2 [/tex]
    [tex] R=v_ocosθ/g * (v_osinθ +√(v_o^2sin^2θ)) [/tex]

    3. The attempt at a solution

    We have to calculate the angle
    [tex] h=(v_o^2sin^2θ)/2g[/tex]

    [tex] 20.4=25^2sin^2θ/2g [/tex]

    [tex] 20.4=31.8878sin^2θ[/tex]

    divide both sides by 31.8878

    [tex] 0.639744 = sin^2θ [/tex]

    take the square root of both sides

    [tex] sinθ = 0.79984 [/tex]

    take the inverse sin of both sides

    [tex] θ = 0.927029 [/tex]

    or in degrees

    [tex] θ = 53.11° [/tex]

    Now that we calculated the angle, we calculate the time to reach the maximum height:

    [tex] t=v_osinθ/g[/tex]
    [tex] =25sin(53.11)/9.8 [/tex]
    [tex] = 2.04 s [/tex]

    Now we calculate the horizontal displacement at its maximum height

    [tex] x=v_ocosθt [/tex]
    [tex] =25cos53.11*2.04=30.61 m[/tex]

    Now we calculate the range

    [tex] R=v_ocosθ/g * (v_osinθ +√(v_o^2sin^2θ)) [/tex]
    [tex] = 25cos53.11/9.8 * (25sin53.11+√(25^2sin^253.11)) [/tex]
    [tex] = 61.24 m [/tex]

    Horizontal distance from the wall = [tex]61.24-30.61 = 30.63 m[/tex]

    Is this correct? and if it's not tell me where

    3. The attempt at a solution
  2. jcsd
  3. Dec 20, 2012 #2


    User Avatar
    Homework Helper

    I think you also need to know how far the wall is from the catapult - call that [itex]x_1[/itex]. Your approach assumes that the wall is positioned where maximum height is attained (ie, at half the range) and that the trajectory only just clears it, in which case your answer is correct.

    For this problem it's probably best to eliminate [itex]t[/itex] and work with
    [tex]y(x) = x \tan\theta - \frac{gx^2}{2v_0^2 \cos^2\theta}[/tex]

    Since horizontal distance is maximal when [itex]\theta = \frac14 \pi[/itex], the only question is whether that clears the wall, ie whether
    x_1\tan(\pi/4) - \frac{gx_1^2}{2v_0^2\cos^2(\pi/4)} \geq 20.4.
    If not you'll need to take the smallest [itex]\pi/4 < \theta < \pi/2[/itex] for which the trajectory clears the wall.

    Your formula for the range can be simplified: [itex]0 < \theta < \pi/2[/itex] so [itex]\sin \theta \geq 0[/itex] and [itex]v_0 > 0[/itex]. Therefore
    [tex]R = (v_0/g)\cos\theta (v_0\sin\theta + \sqrt{v_0^2\sin^2\theta})
    = (v_0/g)\cos\theta (2v_0\sin\theta) = (v_0^2/g)\sin(2\theta)[/tex]
    The distance beyond the wall where the projectile lands is then [itex](v_0^2/g)\sin(2\theta) - x_1[/itex].
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