problem03_74

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

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3.74: In the frame of the hero, the range of the object must be the initial separation plus the amount the enemy has pulled away in that time. Symbolically, x v R v x t v x R 0 E/H 0 E/H 0 + = + = , where E/H v is the velocity of the enemy relative to the hero, t is the time of flight, 0x v is the (constant) x -component of the grenade’s velocity, as measured by the hero, and R is the range of the grenade, also as measured by the hero. Using Eq. (3-29) for R , with 1 2 sin 0 = α and 2 / 0 0 v v x = , . 0 ) 2 ( or , 2 0 0 E/H 2 0 0 E/H 0 2 0 = - - +
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Unformatted text preview: = gx v v v g v v x g v This quadratic is solved for km/h, 1 . 61 ) 4 2 2 ( 2 1 2 E/H E/H = + + = gx v v v where the units for g and x have been properly converted. Relative to the earth, the x-component of velocity is km/h 133.2 45 cos km/h) (61.1 km/h . 90 = ° + , the y-component, the same in both frames, is km/h 43.2 45 sin km/h) 1 . 61 ( = ° , and the magnitude of the velocity is then 140 km/h....
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## This document was uploaded on 02/04/2008.

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