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Unformatted text preview: Physics 325, Fall 2010 Prof. Susan Lamb Homework Assignment #1 Solutions 1) (5 points) For the expansion of f ( x ) = √ 1 + x , knowing f (1) = √ 2, f ′ (1) = 1 2 √ 2 , f ′′ (1) = 1 8 √ 2 therefore, we have √ 1 + x = √ 2 + 1 2 √ 2 ( x 1) 1 16 √ 2 ( x 1) 2 + ... 2 (15 points total) 2a) ( 7 points ) We can find the distance to impact by requiring that the projectile follows its usual parabolic path, and also lands on the straight, sloped line of the hill. Denoting the xaxis to be horizontal and the yaxis to be vertical, the component equations for the impact are: v cos αt = r cos β in the xdirection, and v sin αt 1 2 gt 2 = r sin β in the ydirection, where r is the distance up the hill we’re solving for. The projectile problem without the hill is solved by finding the time of flight first, and we use the same method here. That is, we eliminate t from our equations. It is easier to solve the x equation for t . t = r cos β v cos α We then insert this t into the second equation and, after a small amount of simplification, we get gr 2 cos 2 β 2 v 2 cos 2 α + r sin β r cos β sin α cos α = 0 This is quadratic in r , but it is easy to solve because we already know that r = 0 is a solution because the projectile is on the ground when it is launched. We are interested in the other solution r > 0. As r is not zero, we can divide by r and solve the remaining linear equation. r = 2 v 2 cos 2 α g cos 2 β parenleftbigg cos β sin α cos α sin β parenrightbigg We can also write this as r = 2 v 2 cos 2 α g cos β (tan α tan β ) After checking the units, you can also check that this solution is right by setting...
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This note was uploaded on 10/06/2011 for the course PHYS 325 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Staff
 mechanics, Work

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