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Projectile Motion - Range of projectile

  1. Feb 26, 2014 #1
    1. The problem statement, all variables and given/known data

    My problem consists of two parts, the first being how far from the top of a cliff will a projectile (fired from a cannon) land with a velocity of 120 ms^-1, fired horizontally from cliff height of 150m.
    The second part is where I am struggling, the cannon is now aimed 45° above horizontal, how far will the projectile travel now?

    Known data for part 1.
    Δy = -150
    u(y) = 0 ms^-1
    a(y) = -9.8 ms^-2
    u(x) = 120 ms^-1


    2. Relevant equations

    Δy = u(y)*t + 1/2*a(y)*t^2
    Δx = u(x)*t

    x any y component equations.
    u(x) = u*cos(θ)
    u(y) = u*sin(θ)

    3. The attempt at a solution

    Part 1 I found quite simple so hopefully this is correct..

    Δy = u(y)*t + 1/2*a(y)*t^2
    -150 = 0*t + 1/2(-9.8)*t^2 (substitute above data)
    -150 = -4.9*t^2
    t^2 = -150/-4.9
    t = 5.53 seconds

    Δx = u(x)*t
    Δx = 120*5.53
    Δx = 663.9 m

    Part 2.

    u(x) = 120*cos(45)
    u(x) = 84.85 ms^-1 (2 d.p.)

    u(y) = 120*sin(45)
    u(y) = 84.85 ms^-1

    This is as far I got (finding the x and y components of the initial velocity).
    My issue is that I am not sure how to account for the fact that the cannon is elevated 150 m along with the angle change from horizontal to 45°.
    Any help will be greatly appreciated.
     
  2. jcsd
  3. Feb 26, 2014 #2
    The equation you used in Part 1 for ##\Delta y## works in Part 2, too. Except that ##u_y## is not zero in Part 2 (as you have already found).
     
  4. Feb 26, 2014 #3
    Which of these should the equation look like?
    -150 = 84.85*t + 1/2(-9.8)*t^2
    or
    Δy = 84.85*5.53 + 1/2(-9.8)*(5.53)^2

    With the first I have an issue because I am unable to find a value for t and my value for Δy (-150) will also be incorrect as the projectile is now being fired at an angle.
    With the second I am using an incorrect value for t as this was the value used for the horizontally fired projectile.
     
  5. Feb 27, 2014 #4

    ehild

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    Homework Helper
    Gold Member

    The cannon is on the top of the hill at hight of 150 m, and its tube is directed at 45°angle. The velocity of the cannon ball has a positive vertical component so it takes longer time to reach the ground as previously, when it was projected horizontally. Determine the time from the equation -150 = 84.85*t + 1/2(-9.8)*t^2. As the cannon is small with respect to the hill you can take Δy=-150 m again.

    ehild
     
  6. Feb 27, 2014 #5
    Firing at the angle does not change the vertical displacement. Why are you unable to find a value for t? It is a quadratic equation.
     
  7. Feb 27, 2014 #6
    Of course, I hadn't thought of quadratics (I am just learning kinematics via correspondence). Thank you for your help!
     
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