Javelin Handout Ma123 F05

Calculus (With Analytic Geometry)(8th edition)

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Finding the optimal angle to throw the javelin (p. 185 Calculus text) Finding the optimal angle at which to throw the javelin: 2 2 2 0 sec (tan ) , 0 2 2 g y x x h v θ π = + + Verification of this equation relating x and y : If 0 ν is the velocity at an angle of θ with the horizontal, then 0 cos ν θ is the velocity in the horizontal direction and 0 sin ν θ is the velocity in the vertical direction. The above equation will be verified in class using the following two facts (good, to a first approximation, assuming no air resistance): 1. The acceleration due to gravity is constant (and is equal to g ) and only affects the motion in the vertical direction. 2. The velocity in the horizontal direction is constant (until the javelin hits the ground). Now, setting 0 y = we find that the (positive) solution for the distance the javelin travels is given by 2 2 0 0 0 cos sin sin 2 v x v v gh g = - - - (*) Computing the derivative of x with respect to we find 2 2 2 0 2 2 0 2 0 2 cos sin sin sin 2 sin v dx gh d g v gh v - = - + - + - Since the left-most term in square brackets is positive (note that
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This note was uploaded on 01/22/2008 for the course MATH 123 taught by Professor Johnson during the Fall '06 term at SDSMT.

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Javelin Handout Ma123 F05 - Finding the optimal angle to...

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