Recent content by prelon

A very banal mistake indeed. That was slightly embarrassing. With that, the rest of the solution should look like this: ##sin(2\phi)cos(\theta)-2sin^2(\phi)sin(\theta)## ##=sin(2\phi)cos(\theta)-cos(2\phi)sin(\theta)-sin(\theta)##...

I think I am starting to get what is happening. Most likely I am making a very banal mistake which is making this a lot more complicated for me. I will try to explain my thinking and maybe you will be able to see where I am wrong. The way I am thinking, to use the ##cos(2a) = 1-2sin^2(a)## I...

I have been pondering over this the last few days, but without much result. I have ##R= \frac{v_0{}^2}{g} \frac{sin(2\phi)cos(\theta)-2sin^2(\phi)sin(\theta)} {cos^2(\theta)}## I need to convert the ##2sin^2(\phi)## into ##cos(2\phi)## using ##cos(2a) = cos^2(a)-sin^2(a) = 2cos^2(a)-1 =...

Sorry for taking so long to respond. I am guessing that you are referring to ##sin(2a) = 2sin(a)cos(a)## and ##cos(2a) = cos^2(a)-sin^2(a) = 2cos^2(a)-1 = 1-2sin^2(a)## Using those I get ##R = \frac{2v_0{}^2}{g} \frac{cos(\theta + \phi)sin(\phi)}{cos^2(\theta)} \leftrightarrow R =...

Hmm… you mean writing the function for R so the functions for ##\phi## becomes functions for ##2\phi## (just making sure I understand) In doing so, would I be right in focusing on ##sin(\phi)cos(\phi+\theta)## and ignoring ##\frac{2v_0{}^2}{g cos^2\theta}## ?

I don't think I understand it. If I try to ignore the ##R = \frac{2v_0{}^2}{g cos^2(\theta)}## part I get the following: ##R = sin(\phi)cos(\phi+\theta)## I can then divide that into two functions: ##f(\phi) = sin(\phi)## and ##g(\phi) = cos(\phi+\theta) =...

I am confused. What is a and b, and what is it the derivative of?

Thanks for the reply. I had forgotten all about the rules for taking the derivative of multiple functions. Using the rule you linked to I got ##\frac{dR}{d\phi} = \frac{2v_0{}^2}{g}...

Thanks for the reply. I believe that is what I did in part a). I resolved g and ##v_0## into their x- and y-components along the incline and perpendicular to the incline. I got ##x = v_0cos(\phi)t-0,5gsin(\theta)t^2## and ##y = v_0sin(\phi)t-0,5gcos(\theta)t^2## Then I found that ##t = 0## or...

1. Homework Statement (The following is taken from Sears and Zemansky’s University physics with modern physics, thirteenth edition by Young Freedman. Chapter 3 bridging problem: Launching up an incline at page 95. You fire a ball with an initial speed ##v_0## at an angle Φ above the surface...