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W3.2 - Physics for Scientists Engineers 1 Spring Semester...

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January 18, 2008 Physics for Scientists&Engineers 1 1 Physics for Scientists & Physics for Scientists & Engineers 1 Engineers 1 Spring Semester 2008 Lecture 10
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January 18, 2008 Physics for Scientists&Engineers 1 2 Review Review Projectile trajectory Maximum height reached at Range: Maximum range: Constant velocity along x x H = v 0 2 2 g sin2 ! 0 H = y 0 + v y 0 2 2 g R = v 0 2 g sin2 ! 0 = 2 x H R = v 0 2 g
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January 18, 2008 Physics for Scientists&Engineers 1 3 Example Baseball: Batting Example Baseball: Batting Question: If ball comes off bat with launch angle of 35° and with initial speed of 110 mph, how far will it fly? How long will it be in the air? What will be its speed at the top of its trajectory? What will be its speed when it lands? Answer: (neglect air resistance for now, return to this later) Convert to SI: v 0 = 110 mph = 49.2 m/s. Range: Air time: R = v 0 2 g sin2 ! 0 = (49.2 m/s) 2 9.81 m/s 2 sin(70 ° ) = 231.5 m = 760 ft t = R v 0 cos ! 0 = 231.5 m (49.2 m) " cos(35 ° ) = 5.74 s
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January 18, 2008 Physics for Scientists&Engineers 1 4 Example Baseball: Batting (2) Example Baseball: Batting (2) Answer (cont.): At the top of the trajectory velocity has only a horizontal component : Speed when baseball lands is the same as the speed when it left the bat, here 49.2 m/s
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