The time of landing is t 2 y0 g and the x coordinate

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Unformatted text preview: v0 cosθ0, in the positive x direction. Let M be the mass of the shell and let V0 be the velocity of the fragment. Then Mv0cosθ0 = MV0/2, since the mass of the fragment is M/2. This means V0 = 2 v0 cosθ 0 = 2 ( 20 m/s ) cos 60 ° = 20 m/s. This information is used in the form of initial conditions for a projectile motion problem to determine where the fragment lands. Resetting our clock, we now analyze a projectile launched horizontally at time t = 0 with a speed of 20 m/s from a location having coordinates x0 = 17.7 m, y0 = 15.3 m. Its y coordinate is given by 1 y = y0 − 2 gt 2 , and when it lands this is zero....
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This document was uploaded on 03/24/2014.

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