# 121_2 - CHAPTER 2 PHYS 121 WEEK 2 APRIL 9-13 SPRING 2007...

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Unformatted text preview: CHAPTER 2 PHYS 121 WEEK 2 APRIL 9-13 SPRING 2007 KINEMATICS: DESCRIPTION OF MOTION 93. (a) We take y o = 0. y = y o + v o t − 1 2 g t 2 = − 1 2 g t 2 . So y is proportional to the time squared. Therefore twice the time means (3) four times the height. Given: v o = 0, t = 1.80 s. Find: y A and y B . (b) y A = − 1 2 (9.80 m/s 2 )(1.80 s) 2 = − 15.9 m. So the height of cliff A above the water is 15.88 m = 15.9 m . y B = y A 4 = 15.88 m 4 = 3.97 m . 95. Given: v o = 0, y = − 0.157 m (take y o = 0). Find: t . y − y o = v o t − 1 2 g t 2 = − 1 2 g t 2 , t = 2 y − g = 2( − 0.157 m) − 9.80 m/s 2 = 0.18 s < 0.20 s. It takes less than the average human reaction time for the dollar bill to fall. So the answer is no , not a good deal . 98. The maximum initial velocity corresponds to the apple reaching maximum height just below the ceiling. Given: v = 0 (max height), ( y − y o ) = 3.75 m – 0.50 m = 3.25 m. Find: v o . v 2 = v 2 o – 2 g ( y − y o ), v o = v 2 + 2 g ( y − y o ) = 0 + 2(9.80 m/s 2 )(3.25 m) = 7.98 m/s. Therefore it is slightly less than 8.0 m/s . 103. (a) Given: v o = 0, t = 1.26 s, ( y − y o ) = − 1.30 m (downward). Find: g . y = y o + v o t – 1 2 gt 2 , − 1.30 m = 0 − 1 2 g (1.26 s) 2 . Solving, g = 1.64 m/s 2 . (b) v = v o – gt = 0 − (1.64 m/s 2 )(1.26 s) = − 20.7 m/s = 2.07 m/s downward....
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## This note was uploaded on 04/10/2008 for the course PHYS 121 taught by Professor Zammit during the Spring '07 term at Cal Poly.

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121_2 - CHAPTER 2 PHYS 121 WEEK 2 APRIL 9-13 SPRING 2007...

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