Ballistic pendulum

Feb 3rd, 2012
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The document illustrates how to find the speed of ejection ball by using the projectile motion equations and the angular speed of the pendulum by using the conservation of momentum. Every step is explained in great details.

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Ballistic Pendulum1. What is the ejection speed of a ball if it was shot horizontally 1 meter abovethe floor, and touched it 2.5 meters further from the vertical projection of theejection point? Use 9.810 m/s^2 for the acceleration of gravity.This is a projectile motion problem.The 4 basic equations are:y = y0+V0y t -1/2 gt2Vy = V0y gtx = x0 + V0xtVx = V0xFrom the problem, we have y0 = 1 m, y=0, V0y =0, x0 = 0, and x = 2.5 m, g=9.81m/s2, we need to find vx0, the third equation is the one we want to use, but weneed t to find V0x, that can be found from the first equationy = y0+V0y t -1/2 gt20 = 1 -1/2 gt2t2 = 2/gt=2= 0.4515 sgthen use the third eq,x = x0 + V0xtV0x = x0/t = 5.537 m/s2. Identical shot from the apparatus of Problem 1 was caught by the pendulumbasket initially at rest. From angular momentum conservation, find the angularspeed of the system "pe

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