Dynamics_Part46 - z t f = h wherefore h = eV sin θ w eV g...

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Next, we use the fact that in order for the velocity vector to exactly horizontal, we must have v I z = ˙ z ( t f )=0 . Thus, we have ˙ z ( t f )= eV sin θ gt f =0 = eV sin θ = gL V cos θ Solving for L , there follows L = e V 2 g sin θ cos θ It is helpful to combine the equations for t f and L , which yields t f = eV g sin θ The ball’s height when it reaches the upper surface is
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Unformatted text preview: z ( t f ) = h , wherefore h = eV sin θ w eV g sin θ W − 1 2 g w eV g sin θ W 2 Simplifying and combining like terms, the solution for h is h = 1 2 e 2 V 2 g sin 2 θ...
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This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.

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