Problem Set 1 Solutions

Problem Set 1 - Physics 200 Problem Set 1 Solution Note It's not very fun to punch numbers into a calculator Plugging in numbers at the very end

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Physics 200 Problem Set 1 Solution Note: It’s not very fun to punch numbers into a calculator. Plugging in numbers at the very end will often save you time and mistakes. This won’t matter so much in this problem set, but try to get in the habit now. 1. From the top of a building of height h = 100 m I throw a stone up with velocity 10 m/s. What is the maximum height it reaches, and when does this occur? How many seconds does it spend on its way down between h = 50 m and h = 0 m? What is its velocity when h = 50 m? If, while the stone is airborne, an earthquake opens up a hole 50 m deep in the ground, when and with what speed will the stone hit the bottom? Answer: When the stone reaches its maximum height its velocity is momentarily zero. So 0 = v 0 - gt = t = v 0 g = 10 m / s 9 . 81 m/s 2 = 1 . 0 s , and 0 = v 2 0 - 2 g Δ h = Δ h = v 2 0 2 g = (10 m / s) 2 2(9 . 81 m/s 2 ) = 5 . 1 m . This means the actual height reached by the stone is h building + Δ h = 105 . 1 m . When the stone is at some height Δ h measured from the top of the building its velocity is v 2 = v 2 0 - 2 g Δ h, and between two heights Δ h 1 and Δ h 2 v 2 = v 1 - gt = t = v 1 - v 2 g . Plugging in the numbers gives (negative because the stone is moving downward) v 1 = - q v 2 0 - 2 g Δ h 1 = - q (10 m / s) 2 + 2(9 . 81 m/s 2 )(50 m) = - 32 . 9 m / s , v 2 = - q v 2 0 - 2 g Δ h 2 = - q (10 m / s) 2 + 2(9 . 81 m/s 2 )(100 m) = - 45 . 4 m / s and t = - 32 . 9 m / s - ( - 45 . 4 m / s) 9 . 81 m/s 2 = 1 . 3 s . The velocity when
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This note was uploaded on 07/13/2008 for the course PHYS 200 taught by Professor Ramamurtishankar during the Fall '08 term at Yale.

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Problem Set 1 - Physics 200 Problem Set 1 Solution Note It's not very fun to punch numbers into a calculator Plugging in numbers at the very end

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