October 18th, 2012, 11:33 AM  #1 
Member Joined: Dec 2010 Posts: 51 Thanks: 0  max length of trajectory
What is the angle needed to launch an object from ground level to achieve the maximum length of trajectory? The initial velocity is 96 feet per second. The answer is 56.5 degrees. Here's what i have: http://i.imgur.com/zsRxu.jpg [attachment=0:1o0mkv9k]zsRxu.jpg[/attachment:1o0mkv9k] When i use the graphing calculator to graph , the graph is undefined at 56.5 degrees. What am i doing wrong? The solution manual has this equation (or something like it): which is probably Edit: I used the ti89 to evaluate the integral. Edit2: Oh wow, lol nevermind. i got it now. I had a negative for my 1024t^2 instead of a positive. 
October 18th, 2012, 10:22 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs  Re: max length of trajectory
Eliminating the parameter , we find the object's trajectory will be given by: At this point I wanted to be slick and use the derivative form of the FTOC to differentiate directly, but I couldn't make it work. So, here goes... Let: and we have: And now we may write: If we observe that this integral is symmetric about the line we may write: Let: and we have: Let: and we have: Let: and we have: With a bit of algebra, we may write this in a form we may integrate directly: and so we have (given ): Now, differentiating and equating to zero, we find: And so we find we want to satisfy: Using Newton's method, we will need: And so Newton's method gives us the recursion: Using as the first guess, we find: Converting from radians to degrees, we find the launch angle which maximizes the arc length is: 
October 19th, 2012, 11:58 PM  #3 
Member Joined: Dec 2010 Posts: 51 Thanks: 0  Re: max length of trajectory
WOW! That's some impressive mathematical calculations. I'll saved this and exam it more carefully once my workload from ODE and calc3 clears.


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