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PHY183-Lecture10

# PHY183-Lecture10 - Review Yesterday Projectile trajectory...

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January 26, 2006 Physics for Scientists&Engineers 1 1 Physics for Scientists & Physics for Scientists & Engineers 1 Engineers 1 Spring Semester 2006 Lecture 10 January 26, 2006 Physics for Scientists&Engineers 1 2 Review: Yesterday Review: Yesterday ! 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 January 26, 2006 Physics for Scientists&Engineers 1 3 Example Baseball: Batting Question: ! 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) ! 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 January 26, 2006 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 Remember that in general And since y=y 0

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