Physics 325 Homework Solutions

# Physics 325 Homework Solutions - Physics 325 Fall 2010 Prof...

This preview shows pages 1–2. Sign up to view the full content.

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 x -axis to be horizontal and the y -axis to be vertical, the component equations for the impact are: v 0 cos αt = r cos β in the x-direction, and v 0 sin αt - 1 2 gt 2 = r sin β in the y -direction, 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 0 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 0 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 0 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 0 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 β = 0 and checking that r ( β = 0) =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern