Exam1-Solutions-S14

The change in its velocity is 1 500 ms south 2 600

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: can be o solved to give ymax = v2 o . We have two cases with different initial velocities. 2g v 21 v 22 o For case 1, ymax 1 = for case 2, ymax 2 = o . Dividing the first equation by 2g 2g the second gives ymax 1 ymax 2 v 21 o 2 v 21 2g v o1 o = 2 = 2 . So ymax 1 = 2 ymax 2 v o2 v o2 v o2 2g PHY 2048 Spring 2014 – Acosta, Rinzler Exam 1 solutions 2 ⎛v ⎞ Hence the height of the first will be Q = 2 = ⎜ o1 ⎟ times the height of the 2nd. v o2 ⎝ v o2 ⎠ Since the speed with which the 1st was thrown was 100 m/s and the speed of the v 21 o 2 ⎛ 100 ⎞ 2nd was 10 m/s , Q = ⎜ = 100. ⎝ 10 ⎟ ⎠ Problem 9 A ball is thrown horizontally from the top of a cliff from a height h=23.0 m (or 32 m, 41 m) above the level plain below. It strikes the plain at an angle of 60° with respect to the horizontal. With what...
View Full 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