My Math Forum Path of Projectile

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 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=(10x-x^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
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Quote:
 Originally Posted by waz 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?
He means take the derivative of y(x), which gives y'(x) and then solve the equation y'(x) = 0.

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

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