problem03_15

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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3.15: a) Solving Eq. (3.17) for 0 = y v , with ° = 0 . 45 sin ) s m 0 . 15 ( 0 y v , s. 08 . 1 s m 80 . 9 45 sin ) s m 0 . 15 ( 2 = ° = T b) Using Equations (3.20) and (3.21) gives at m) 52 . 4 , m 18 . 6 ( ) , ( , 1 = y x t : ) m 52 . 4 , m 8 . 16 ( , : ) m 74 . 5 , m 5 . 11 ( , 3 2 t t . c) Using Equations (3.22) and (3.23) gives at ), s m 9 . 4 , s m 6 . 10 ( : ) 0 , s m 6 . 10 ( , : ) s m 9 . 4 , s m 6 . 10 ( ) , ( , 3 2 1 - = t t v v t y x for velocities, respectively, of s m 7 . 11 @ 24.8 ° , s m 6 . 10 @ 0 ° and s m 7 . 11 @ - 24.8 ° . Note that x v is the same for all times, and that the y -component of velocity at 3 t is negative that at 1 t . d) The parallel and perpendicular components of the acceleration are obtained from . , , ) ( || || 2 || a a a v a a v v a a v v v & - = = = v v For projectile motion, y gv g - = - = v a j a so , ˆ , and the components of acceleration
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Unformatted text preview: parallel and perpendicular to the velocity are 2 2 1 s m 9 . 8 , s m 1 . 4 :-t . 2 2 s m 8 . 9 , : t . 2 2 3 s m 9 . 8 , s m 1 . 4 : t . e) f) At t 1 , the projectile is moving upward but slowing down; at t 2 the motion is instantaneously horizontal, but the vertical component of velocity is decreasing; at t 3 , the projectile is falling down and its speed is increasing. The horizontal component of velocity is constant....
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