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Projectile motion

  1. Oct 12, 2015 #1
    I've put the problem statement below and worked it out. I typically don't post questions like this as they're a lot to go through, but I am wondering if I have worked the problem correctly as my book does not have the solution and I feel like I am not understand the material correctly.

    1. The problem statement, all variables and given/known data


    A shell is fired upward from the top of a building at an angle of [itex]\pi /3[/itex] with the horizontal. Its initial velocity is 60 m/s, and assume it is fired from the point (0, 500) when [itex]t=0[/itex] and [itex]x''(t) = 0[/itex] and [itex]y''(t)=-9.8 m/s^2[/itex] while [itex] 0 \le t \le t_2[/itex] is the time when the shell hits the ground.

    3. The attempt at a solution

    Given that [itex]x''(t) = 0[/itex] and [itex]y''(t) = -9.8[/itex] we have r''(t) = <0, -9.8> integrating the vector-valued function we have [tex]r'(t) = <C, -9.8t + C> \rightarrow r'(t) = <60cos(\pi/3), -9.8t + 60sin(\pi/3)>[/tex] using the initial velocity given. Integrating again to get the position we have [tex] r(t) = <60cos(\pi/3)t + C, -4.9t^2 + 60sin(\pi/3)t + C>[/tex] [tex]r(t) = <60cos(\pi/3)t, -4.9t^2 + 60sin(\pi/3)t + 500>[/tex] using the point (0, 500) given.

    Now, are the parametric equations [itex]x(t)[/itex] and [itex]y(t)[/itex] for [itex]r(t)=<x(t),y(t)>[/itex] just [itex]x(t) = 60cos(\pi/3)t[/itex] and [itex]y(t) = -4.9t^2 + 60sin(\pi/3) + 500[/itex]?

    Solving for the time the shell reaches the maximum height will be the time at which [itex]y'(t) = 0[/itex]? so we have: [tex]-9.8t + 60sin(\pi/3) = 0 \Rightarrow t = 60sin(\pi/3)/9.8 = 5.3 s[/tex]
    The maximum height will then be [tex] y(5.3) = -4.9(5.3)^2 + 60sin(\pi/3)(5.3) + 500 = 637.75 m[/tex]
    The time the shell hits the ground is when [itex]y(t) = 0[/itex]? so we have: [tex] 0 = -4.9t^2 + 60sin(\pi/3)t + 500 \Rightarrow t = 10.11[/tex]
    Horizontal distance: [itex]x(10.11) = 60cos(\pi/3)(10.11) = 303.3 m[/itex]

    The speed of shell at impact:

    [itex] ||r'(10.11)|| = \sqrt{(60cos(\pi/3))^2 + (-9.8(10.11) + 60sin(\pi/3))^2} = 55.86 m/s[/itex]

    Is this the correct way to calculate the speed? As speed is just the magnitude of the velocity?
     
    Last edited: Oct 12, 2015
  2. jcsd
  3. Oct 12, 2015 #2
    The angles are different. Am I missing something?
     
  4. Oct 12, 2015 #3
    No, sorry I mistyped I'll fix it.
     
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