ch05-p064 - 64. We first use Eq. 4-26 to solve for the...

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64. We first use Eq. 4-26 to solve for the launch speed of the shot: 2 0 2 (tan ) . 2( cos ) gx yy x v θ −= With 34.10 , 0 2.11 m y = and ( , ) (15.90 m,0) xy = , we find the launch speed to be 11.85 m/s. v ′ = During this phase, the acceleration is 22 2 0 (11.85 m/s) (2.50 m/s) 40.63 m/s . ( 1 . 6 5 m ) vv a L ′ − == = Since the acceleration along the slanted path depends on only the force components along
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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