is the initial speed and θ0is the firing angle. The coordinates of the highest point on the trajectory are ()220000220 m/scossincossin 60 cos6017.7 m9.8 m/sxvxvtvtgθθ====°°=andyvt gtvgy=−===°02022022121212209860153sin..m / sm/sm.2bgSince 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 v0cos0, in the positive xdirection. Let Mbe the mass of the shell and let V0be the velocity of the fragment. Then Mv0cos0= MV0/2, since the mass of the fragment is M/2. This means Vv00 022 206020=°coscosm/s.bgThis 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.