HW6-phy2048-Spring-2014-solution

# Since one fragment has a velocity of zero after the

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Unformatted text preview: is zero at time t = v0 y/g = (v0/g) sin θ0, where v0 is the initial speed and θ0 is the firing angle. The coordinates of the highest point on the trajectory are 2 ( 20 m/s ) sin 60° cos 60° = 17.7 m v0 x = v0 xt = v0t cos θ0 = sin θ0 cos θ0 = g 9.8 m/s 2 2 and y = v0 y t − b g 2 2 1 2 1 v0 1 20 m / s gt = sin 2 θ 0 = sin 2 60° = 15.3 m. 2 2 2g 2 9.8 m / s 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 1 PHY2048 Spring 2014 HW6 the highest point the velocity of the shell is...
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