position for this part of the problem) and take +xrightward and +ydownward. Then (since the initial velocity is purely horizontal) the projectile motion equations become xvtygt==−and122.Solving for xat the time when y = h, the second equation gives thg=2.Then, substituting this into the first equation, we find ()22 2.0 m27.48 m/s4.8 m.9.8 m/shg=11. To find where the ball lands, we need to know its speed as it leaves the track (using conservation of energy). Its initial kinetic energy is Ki= 0 and its initial potential energy isUi= M gH. Its final kinetic energy (as it leaves the track) isKMvIf=+122122ω(Eq. 11-5) and its final potential energy is M gh. Here we use vto denote the speed of its
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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.