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Kinematics problem solving

  1. Jul 9, 2011 #1
    1. The problem statement, all variables and given/known data

    A cannon which is at the top of a 99 m high cliff fires in a horizontal direction to a dog. The initial position of the dog is 1000 m from the cliff's base. What should be the initial velocity of the cannon if the dog is moving at constant speed of 9 m/s to the cliff.

    Other given
    square root of (198 m/9.8 m/s) = 4.49 s

    2. Relevant equations

    Yf = Yi + Vi(t) + (1/2)(gt^2)
    Xf = Xi + Vi(t) + (1/2)(gt^2)

    3. The attempt at a solution

    For the y-component:

    99 m = 0 + Vi(4.49 s) + (1/2)(-9.8 m/s^2)(4.49)^2
    = Vi(4.49 s) -99.78 m - 99 m
    = Vi(4.49 s) - 198.78 m

    For the x-component:

    0 m = 1000 m + (-9 m/s)(4.49 s) + (1/2)(-9.8 m/s^2)(4.5)^2
    = 860.365 m


    Xi - Xf + Vi(t) + (1/2)(gt^2) = Yi - Yf + Vi(t) + (1/2)(gt^2)
    Vi(4.49 s) - 198.78 m = 860.365 m
    Vi(4.49 s) = 860.365 + 198.78 m
    Vi = 1059.145 m/4.49 s
    Vi = 235.89 m/s

    Can someone verify this? Thanks alot.
  2. jcsd
  3. Jul 9, 2011 #2


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    Could you explain what your notations mean?

  4. Jul 9, 2011 #3
    Yf = Yi + Vi(t) + (1/2)(gt^2)
    Xf = Xi + Vi(t) + (1/2)(gt^2)

    Yf = final position (y-axis)
    Xf = final position (x-axis)
    Vi = initial velocity
    t = time
    g = gravitational pull
  5. Jul 9, 2011 #4


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    Final and initial position of what? There are two moving things: the cannon ball and the dog. They start from different initial positions and they meet, so the final positions are the same for both of them. The ball accelerates downward, but the dog moves with constant velocity.

  6. Jul 9, 2011 #5
    actually, I'm not sure about the equations that I used.
  7. Jul 9, 2011 #6


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    There are the Yi and Yf and Xi and Xf data for the ball, and xi and xf for the dog. As it moves on the ground, yi=yf=0. The position of both the ball and the dog are the same when the ball hits the dog, so Yf =0 and Xf=xf. You calculated the time needed to the ball to reach the ground, t=4.49 s. During that time the dog walked some distance, with constant velocity v= 9m/s towrds the cliff. So how far is it from the cliff when the ball hits the ground? The ball has to reach there at the same time as the dog.

    You have to know that the motion of a projectile can be treated as two separate motions, one vertical with the downward acceleration -g, the other horizontal with constant velocity Vi. Instead of your equation, Xf=Xi +Vi t. But we measure the distance from the rock, so Xi=0, and Xf is the same as the distance of the dog from the cliff at time 4.49 s. Find Vi.

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