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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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?

Geometry
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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
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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
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
* typo it should be y = -0.25(x - 20)^2 + 100
the distance from the base would be the value of x when y = -125
-125 = -0.25(x - 20)^2 + 100 (x - 20)^2 = -125-100 / -0.25 = 900 x - 20 = 30
so distance from base is 50 m

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