ch05-p049 - 49. Using Eq. 4-26, the launch speed of the...

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49. Using Eq. 4-26, the launch speed of the projectile is 2 0 (9.8 m/s )(69 m) 26.52 m/s sin 2 sin 2(53 ) gR v θ == = ° . The horizontal and vertical components of the speed are 0 0 cos (26.52 m/s)cos53 15.96 m/s sin (26.52 m/s)sin53 21.18 m/s. x y vv ° = ° = Since the acceleration is constant, we can use Eq. 2-16 to analyze the motion. The component of the acceleration in the horizontal direction is 2 2 2 (15.96 m/s) 40.7 m/s , 2 2(5.2 m)cos53 x x v a x = ° and the force component is
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