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SampleTestSoln2Q13

# SampleTestSoln2Q13 - 2-13 Take the origin as being the...

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2-13 Take the origin as being the original point of release of the arrow from the bow. Thus we know that ! y = - 18 [m]. Because there is no acceleration in the x direction, we know that the x component of the velocity of the arrow is unchanged during the motion. Thus v x = v 0x = 55 [m/s] cos 25 = 49.8 [m/s]. The y component of the final velocity of the arrow, v y = - 55 [m/s] sin 25 = -23.2 [m/s] with the – sign signifying that the arrow’s velocity is downward. Now we can use the relation v y 2 = v 0y 2 + 2 a ! y Thus v 0y = " { v y ) 2
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