chp2 problem57 soln

Physics: Principles with Applications

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View Full Document Right Arrow Icon Solution (a) To solve this problem we will break the fall into two parts. In part (I) we will consider the first 15 meters of the fall to the safety net. In part (II) we will calculate the deceleration caused by the net stretching. Part (I) In part (II) we will need the velocity of the person when they hit the net which is the velocity after the 15 meters of free fall. Considering the first 15 meters and letting down be positive then. We can write Eq. 2-10c as Since the distance the persons falls is 15 meters Plugging in we get Part (II) Now consider the person's motion after hitting the net.
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Unformatted text preview: Where in part (I) we found the initial velocity. Solving Eq.2-10c for acceleration and plugging in we get The negative sign means the acceleration is upwards which decelerates the person. Solution (b) We see from Eq. 2-10c (1 of 2) [1/31/2008 12:18:18 PM] A larger net stretch ( ) gives a smaller deceleration. So the net should be (2 of 2) [1/31/2008 12:18:18 PM]...
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