ch09-p015

# ch09-p015 - 15 We need to find the coordinates of the point...

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is the initial speed and θ 0 is the firing angle. The coordinates of the highest point on the trajectory are () 2 2 0 00 0 2 20 m/s cos sin cos sin 60 cos60 17.7 m 9.8 m/s x v xv tv t g θθ == = = ° ° = and yv t g t v g y =− = = = ° 0 2 0 2 2 0 2 2 1 2 1 2 1 2 20 98 60 153 sin . . m / s m/s m. 2 b g Since no horizontal forces act, the horizontal component of the momentum is conserved. Since one fragment has a velocity of zero after the explosion, the momentum of the other equals the momentum of the shell before the explosion. At the highest point the velocity of the shell is v 0 cos 0 , in the positive x direction. Let M be the mass of the shell and let V 0 be the velocity of the fragment. Then Mv 0 cos 0 = MV 0 /2, since the mass of the fragment is M /2. This means Vv 00 0 2 2 20 60 20 = ° cos cos m/s. bg 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
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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