SampleTestSoln2Q8-12 - 2-8 There are many ways to solve the...

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Unformatted text preview: 2-8 There are many ways to solve the problem. One way is to use symmetry, in which we would see that the projectile will strike the surface with its downward velocity equal in magnitude to the initial upward component of the velocity. For the vertical component, we'd write vfy = v0y + at and use symmetry to get v0y = v0y + at. With a= -g , we get 2 v0y = gt or t = 2 v0y/g = 2 v0 sin! /g = 29.8 [m/s] (0.5) / 9.8 [m/s2] = 1 s. You can also ask the question how long would it take for the object to come back to the same height !y = 0? Thus !y = 0 = v0yt gt2. Solving we get the solution t=0 or t = v0y / ( g) = 9.8 [m/s] (0.5) / ((0.5) 9.8 [m/s2]) = 1 s. Thus (B) is the correct answer. ...
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This note was uploaded on 09/08/2009 for the course PHYS 1101 taught by Professor Richardson, b during the Fall '08 term at Cornell University (Engineering School).

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