{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw3_solutions - Physics 325 Fall 2006 Prof Susan Lamb...

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

View Full Document Right Arrow Icon
Physics 325, Fall 2006 Prof. Susan Lamb Homework Assignment #3 Solutions 1a) 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 x component equation for the impact is v 0 cos αt = r cos β v 0 sin αt - 1 2 gt 2 = r sin β 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 since we need to eliminate t from our equations, let’s do that here. It’s a lot easier to solve the x equation for t . t = r cos β v 0 cos α We can insert this t into the second equation. 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’s an easy one. We already know r = 0 is a solution since the projectile is on the ground when it is launched. We’re interested in the other solution r > 0, so you can divide by r and solve the remaining linear equation. r = 2 v 2 0 cos 2 α g cos 2 β cos β sin α cos α - sin β 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 is right by setting β = 0 and checking that r ( β = 0) = 2
Background image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}