Dynamics_Part48

Dynamics_Part48 - So the velocity vector just after the...

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So, the velocity vector just after the impact is v I = v I n n + v I t t = eU sin θ ( i sin θ k cos θ )+ U cos θ ( i cos θ + k sin θ ) Regrouping terms, we find v I = U D cos 2 θ e sin 2 θ i i +(1+ e ) U sin θ cos θ k (b) If the ball moves without friction, its vertical motion is governed by m d 2 z dt 2 = mg = z = z o + v oz t 1 2 gt 2 The initial position and velocity are z o = h and v oz =(1+ e ) U sin θ cos θ = 1 2 (1 + e ) U sin 2 θ Therefore, the vertical position of the ball is z = h + 1 2 (1 + e ) Ut sin 2 θ 1 2 gt 2 The ball lands on the girl’s head when z = h ,sotha t h = h + 1 2 (1 + e ) Ut sin 2 θ 1 2 gt 2 = t 2 = (1 + e ) U sin 2 θ g t The solution t =0
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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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