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anonymous

  • one year ago

A cannonball when configured to be fired at a certain angle would have a parabolic path with a maximum height of 100m and a horizontal range of 40m. if the cannonball is placed at the edge of the cliff 125 high, how far from the base would the cannon ball land?

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  1. Jack1
    • one year ago
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    \(\large s = u \times t + \frac 12 \times a \times t^2\) u= initial velocity t = time a = acceleration this is one of the kinematic equations... there are a few, does it look like something you've seen before? you can use this equation and the others, in both the horizontal and vertical planes, to work out the final range of the cannonball

  2. Jack1
    • one year ago
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    |dw:1435499594741:dw|

  3. Jack1
    • one year ago
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    |dw:1435499679676:dw|

  4. anonymous
    • one year ago
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    The problem came from a geometric point of view and not a physics one so I dont really know how to solve it using the eqn you provided

  5. welshfella
    • one year ago
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    we need to work out the equation of the parabola it will be of the form y = -a(x - 20)^2 + 100 as the vertex will be at (20,100) the point (0.0) will also be on the curve 9 the point from which the cannonball was ffired) so we have 0 = -a(0-20)^2 + 100 0 = -400a = 100 a = -0.25 so our equation will be y = -0.25(x - 10)^2 + 100

  6. welshfella
    • one year ago
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    * typo it should be y = -0.25(x - 20)^2 + 100

  7. welshfella
    • one year ago
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    the distance from the base would be the value of x when y = -125

  8. welshfella
    • one year ago
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    -125 = -0.25(x - 20)^2 + 100 (x - 20)^2 = -125-100 / -0.25 = 900 x - 20 = 30

  9. welshfella
    • one year ago
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    so distance from base is 50 m

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