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Solução Do Halliday 8.ed.vol.1. Cap.5 - Ch05 - P064

A solução do Halliday da 8ª edição volume 1. do Cap.5

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    December 2018
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64. We first use Eq. 4-26 to solve for the launch speed of the shot: y − y0 = (tan θ ) x − gx 2 . 2(v′ cos θ ) 2 With θ = 34.10°, y0 = 2.11 m and ( x, y ) = (15.90 m, 0) , we find the launch speed to be v′ = 11.85 m/s. During this phase, the acceleration is a= v′2 − v02 (11.85 m/s)2 − (2.50 m/s)2 = = 40.63 m/s 2 . 2L 2(1.65 m) Since the acceleration along the slanted path depends on only the force components along the path, not the components perpendicular to the path, the average force on the shot during the acceleration phase is F = m(a + g sin θ ) = (7.260 kg) ª¬ 40.63 m/s 2 + (9.80 m/s 2 ) sin 34.10°º¼ = 334.8 N.