May 27th, 2015, 07:10 AM  #1 
Newbie Joined: Nov 2014 From: england Posts: 7 Thanks: 0  Path of Projectile
Hi guys, Below is my question: The path of a projectile is given by: y=(10xx^2)/10 where y is height above the ground and x is distance along the ground from the launch point. Assume the ground to be perfectly flat and horizontal. Distances are measured in metres. Find : (i) the x value at which the projectile reaches maximum height (ii) the maximum height achieved. Any help would be much appreciated! 
May 27th, 2015, 08:01 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
The function you give is a quadratic function with negative leading coefficient so its graph, the trajectory of the projectile, is a parabola opening downward. Its highest point is the "vertex" of the parabola. Since you posted this under "Calculus", there are two ways you can do this. 1) The derivative of height, with respect to x is positive as long as height is increasing, negative when height is decreasing. In between, at the highest point, the derivative is 0. Set the derivative equal to 0 and solve that equation for the time. 2) Complete the square. That allows you to write y= b (x a)^2. Since a square is never negative, this is always b minus something. y will be largest when that "something" is 0 which happens when x= a. 
May 28th, 2015, 01:27 AM  #3 
Newbie Joined: Nov 2014 From: england Posts: 7 Thanks: 0 
I'm not 100% sure i followed all of that Firstly, when you say set the derivative equal to 0 and solve the equation do you mean replace x for 0? 
May 28th, 2015, 02:53 AM  #4  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,882 Thanks: 761 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
For example, let's take the case of a projectile following the path $\displaystyle y(x) = 3x^2 + 6x + 2$. Find the maximum height and where it occurs. The derivative is $\displaystyle y'(x) = 6x + 6$. Solving $\displaystyle y'(x) = 6x + 6 = 0$ gives x = 1. This is where the maximum height is. To find out how high this is we put x= 1 into the formula for y(x) and get $\displaystyle 3(1)^2 + 6(1) + 2 = 5$. I (and others) are happy to help you but you really need to contact your professor or a good tutor to catch you up. Dan  
May 28th, 2015, 06:07 AM  #5 
Senior Member Joined: Jan 2012 From: Erewhon Posts: 245 Thanks: 112 
The parabola has its peak midway between its roots. Here the roots are at $x=0$ and $x=10$, so the maximum height occurs when $x=5$ now find the corresponding $y$. . 

Tags 
calculus, height, path, projectile 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
projectile motion, how to get the equation for range in projectile motion  sweer6  New Users  4  May 21st, 2014 10:00 PM 
projectile motion  shalini maniarasan  Calculus  1  April 25th, 2014 01:26 AM 
path dependent function with a definite path  aise5668  Real Analysis  3  March 5th, 2012 06:36 PM 
Angle of projectile  Canning_s  Algebra  1  November 13th, 2010 11:43 AM 
Projectile Motion  symmetry  Algebra  1  June 19th, 2007 10:26 PM 