ch02-p064 - 64. The height reached by the player is y =...

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2 0 2 2(9.8 m/s )(0.76 m) 3.86 m/s . vg y == = This is a consequence of Eq. 2-16 where velocity v vanishes. As the player reaches y 1 = 0.76 m – 0.15 m = 0.61 m, his speed v 1 satisfies vv g y 0 2 1 2 1 2 −= , which yields 22 2 10 1 2 (3.86 m/s) 2(9.80 m/s )(0.61 m) 1.71 m/s . g y =− = = The time t 1 that the player spends ascending in the top Δ y 1 = 0.15 m of the jump can now be found from Eq. 2-17: () 11 1 1 2 0.15 m 1 0.175 s 2 1.71m/s 0 yv v t t Δ= + ¡ + which means that the total time spent in that top 15 cm (both ascending and descending) is 2(0.175 s) = 0.35 s = 350 ms. (b) The time t 2 when the player reaches a height of 0.15 m is found from Eq. 2-15:
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