# Lecture Notes 8/27

## Physics 114 with Ammar at University of Kansas *

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Lecture Notes August 27th Motion Equations Example a) How High? Given V0=30 m/s & a=g v2 = V02 + 2a∆x 0 = 302 -2g∆x ∆x = 302/-2g ∆x = 45.9 m b) How long does it take to get there? V = V0 + at 0 = 30 - 9.8t t = 3.06 sec. c) Velocity at C? v2 = V02 + 2a∆y V2 = 302 + 2a(0) V=±30 V=-30 m/s d) Velocity at D? v2 = V02 + 2a∆x v2 = 302 - 2g(-50m) v2 = 302 + 100g v = √1880 v= -43.4 m/s e) Time to reach D? ∆X = V0t +½at2 -50 = 30t + ½(-9.8)t2 0=4.9t2 - 30t - 50 Quadratic Equation t = 7.49 seconds Example A car slows uniformly from 22 m/s to rest in 5 seconds. What is the distance traveled? Given: V0 = 22 m/s V = 0 t=5 seconds When acceleration is constant: Vav = V0 + V / 2 V = 22/2 = 11 m/s Example Ballplayer throws a ball up, catches it after 3.3 sec. What velocity did he throw it up with? Given: ∆y=0 t=3.3 s a=-9.8 ∆X = V0t +½at2 0 = V0(3.3) -4.9(3.3)2 4.9(3.3) = V0 V0 = 16.17 m/s

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