calculus to prove angle of maximum range (projectile) (1 Viewer)

[idling fire]

I've been attempting to prove that 45 degrees is the angle required for maximum range of a projectile, but have been unsuccessful in finding the equation that will allow me to differentiate...

It sounded quite simple when I started trying, then an hour later it was still no luck. If I was to make excuses for this, it would be that I haven't actually learnt the formulae for the horizontal/vertical components yet (just read in textbook so very shaky understanding). But still... I thought I should be able to do this...

*tries again* -> nope, I'm lost.
Any hints? Thanks people.

 

[Riviet]

Any hints?

Before you view the hints, see how far as you can before uncovering them. The hints are in sequential order. Also it helps you understand it better if you try it first yourself.

Hint #1:

Spoiler

Starting with
..
y = -g

and

..
x = 0, derive the vertical and horizontal components in terms of t and the angle of projection (say theta)

The most simple scenario would be projecting the object from the origin (ie at x=0 and y=0), this should make your working more simple.

Also the y equations aren't necessary for this proof but you may as well derive them as you will need them for other questions that you encounter in the topic.

Hint #2:

Spoiler

Horizontal range occurs when y=0, find an expression for t and substitute it into the x equation.

Hint #3:

Spoiler

Differentiate x with respect to theta (ie find dx/d@)

Hint #4:

Spoiler

x is a max when dx/d@=0

Hint #5:

Spoiler

Solve the resulting equation from dx/d@=0 to find @

Hope that helps.

 

[airie]

Must you use calculus?

Otherwise you could just work from delta(y)=0 (I'm assuming it is here for your case ), use the formula delta(y) = u_yt + 0.5a_yt², where u_y = u sin(theta), u being the initial velocity and theta the angle this velocity vector makes with the horizontal. Since you know a_y is just gravity ie. -9.8ms^-2, you could solve for t. Then sub into the formula to solve for the range, delta(x) = u_xt, you should be able to prove that theta being 45 degrees would yield the maximum range

Btw, you do need to know that 2cos(theta)sin(theta)=sin(2 theta) though

 

[idling fire]

Otherwise you could just work from delta(y)=0 (I'm assuming it is here for your case ), use the formula delta(y) = u_yt + 0.5a_yt², where u_y = u sin(theta), u being the initial velocity and theta the angle this velocity vector makes with the horizontal. Since you know a_y is just gravity ie. -9.8ms^-2, you could solve for t. Then sub into the formula to solve for the range, delta(x) = u_xt, you should be able to prove that theta being 45 degrees would yield the maximum range

Btw, you do need to know that 2cos(theta)sin(theta)=sin(2 theta) though

I attempted that for a while, but when I reached the part with sin(2 theta) my proof was too messy. I had to use the sine graph to show that sin 90 is max. Not the kind of proof I like, although of course I'm sure there's a way to do it nicely. Grr I'm too stupid to be doing physics >.<

Thanks guys, I'll give Riviet's hints a shot and see what I come up with.

 

[Forbidden.]

In high school Physics, calculus isn't needed, the most you will need to know is to find the gradient. In uni, enginners will need calculus ...

 

[airie]

I attempted that for a while, but when I reached the part with sin(2 theta) my proof was too messy. I had to use the sine graph to show that sin 90 is max. Not the kind of proof I like, although of course I'm sure there's a way to do it nicely. Grr I'm too stupid to be doing physics >.<

It's just...three lines o.0

I think you could just state that between angles 0 to 180 degrees, sin(2 theta) is maximum at 1 when 2 theta is 90 degrees.

But you could do it whatever way to your liking

 

[idling fire]

Hope that helps.

Uhm... differentiate wwith respect to theta? How does that work? Theoretically it sounds ok, but the fact that it's an angle kinda throws me (no projectile pun intended). Is it something like dy/d@ of sin@ = -cos@ ? If so, I don't know how to use it past there, and if not, then I don't have a clue. Yeah, I'm n00b. Not even sure if I got the first hint right. *sigh*

It's just...three lines o.0

Apparently 3 lines is too difficult for myself.

In high school Physics, calculus isn't needed, the most you will need to know is to find the gradient. In uni, enginners will need calculus ...

Unfortunately I'm one of those stubborn people who, if they see a problem and feel like solving it, will continue to attempt to do so until they become physically unable to. (Intellectually unable is not an excuse, even when its beyond my abilities - and in most cases it is.) So back to the drawing board it is.

 

[SoulSearcher]

For sin2@ to be a maximum value, sin2@ = 1, since the range of the sine wave is -1 < x < 1
Therefore
sin 2@ = 1
2@ = 90', since we're looking for an angle between 0' and 90'
Therefore @ = 45'

 

[idling fire]

For sin2@ to be a maximum value, sin2@ = 1, since the range of the sine wave is -1 < x < 1
Therefore
sin 2@ = 1
2@ = 90', since we're looking for an angle between 0' and 90'
Therefore @ = 45'

Hmm yeah messy though (IMO anyway). I dunno, guess I'm just weird. Thanks anyway.