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Unformatted text preview: 213 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 vx = v0x = 55 [m/s] cos 25 = 49.8 [m/s]. The y component of the final velocity of the arrow, vy =  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 vy2 = v0y2 + 2 a!y Thus v0y = " { vy)2  2 (9.8 m/s2) (18 [m])} = " {23.2 [m/s])2  2(9.8 m/s2) (18 [m])} = 13.6 m/s. Note we cannot say whether the arrow was fired with an upward component or a downward component to the vertical velocity. The angle with respect to the horizontal is given by # = tan 1 (vy/vx) = tan1 (13.6 [m/s]) / (49.8 [m/s]) = 15 with respect to the horizontal. The initial speed v = "{(v0x)2+ (v0y)2}= "{(49.8 [m/s])2+ (13.6 [m/s])2}= 52 [m/s]. Thus the correct choice is (C). ...
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This note was uploaded on 09/08/2009 for the course PHYS 1101 taught by Professor Richardson, b during the Fall '08 term at Cornell University (Engineering School).
 Fall '08
 RICHARDSON, B
 Acceleration

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