How far away will the creature be when it falls back down to earth

How far away will the creature be when it falls back down to earth

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How far away will the creature be when it falls back down to earth? Figure %: Diagram of a creature fired out of a canon at an angle θ . To answer this question we must first determine the position function, x (t) , which means we must find v 0 and x 0 . We can choose the x -axis to point in the direction of the creature's horizontal movement across the earth. This means that the creature's movement will be constrained to the x - z plane, and so we can completely ignore the y -direction, effectively reducing our problem to two dimensions. From the initial speed and angle of projection, we can determine that v 0 = (v cosθ, 0, v sinθ) . Since the canon is fired from the surface of the earth, we can set x 0 = 0 (where 0 = (0, 0, 0) , the zero-vector). This leaves us with the position function:
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Unformatted text preview: x (t) = (0, 0, - g)t 2 + (v cosθ, 0, v sinθ)t The y -equation is pretty much useless. If we break this up into x - and z -components we get: x(t) = v cosθt z(t) = v sinθt - gt 2 The next step is to find that time at which the creature will hit the ground. Setting z(t) = 0 and solving for t we find that the time at which the creature will hit the ground is t f = . Finally, we need to plug this time into the equation for the x -position, to see how far the creature has traveled horizontally in this time. x(t f ) = Using the trig identity sin(2θ) = 2 sinθcosθ we find that when the creature hits the ground its distance from the canon will be: x(t f ) =...
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How far away will the creature be when it falls back down to earth

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