Optimum angle of projection (1 Viewer)

[Calculon]

Suppose I get a question like:

a ball can be thrown 50m at an angle of 30 degrees, calculate the optimum range.

Am I allowed to assume that the optimum angle is 45 degrees or do I have to prove it every time?

 

[George W. Bush]

Originally posted by Calculon
a ball can be thrown 50m at an angle of 30 degrees,
optimum angle is 45 degrees

:???:

 

[Zarathustra]

Is this like how the max area of a rectangle is always a square - but you have to prove it.
If these sort of questions were asked in physics I'm sure they'd be on the data sheet - a little something like this

OPTIMUM ANGLE FOR PROJECTILE
45 degrees

 

[Xayma]

Originally posted by George W. Bush
:???:

If it is thrown at 30 degrees it has a range of 50m.

If you were to change the angle what is the maximum distance you could get it. (Basically can you just assume it is 45degrees or do you have to prove it ( and for me: and how do you prove it, I know it is the best just never needed to prove it before))

 

[CrashOveride]

Range = Horziontal velocty x Time of flight

Range = Vcos& . [2Vsin&]/g
Range = [V² 2sin&cos&] /g
Range = [V²sin2&] / g

So the max will be when sin2& = 1, i.e. when & = 45 degrees.

 

[George W. Bush]

ah, i gotcha.

to prove you derive the equation for maximum range (v^2 sin2@ / g if i remember) then state maximum when sin2@ = 1 i.e. @ = 45

 

[Calculon]

I know how, I was just asking if it is necessary.

 

[CrashOveride]

well...i havn't been doing it but maybe in the HSC it might be...im not all to sure about that

 

[Xayma]

Originally posted by Calculon
I know how, I was just asking if it is necessary.

Yeah I asked how, cause I couldn't remember, hmm I think it would be, purely because you have to derive everything else.

 

[CrashOveride]

Derive everything else ?

Like you mean the equations of motion for uniform acceleration?

 

[Calculon]

Originally posted by CrashOveride
Derive everything else ?

Like you mean the equations of motion for uniform acceleration?

yeah

 

[nike33]

the optimal angle in EVERY case is not 45%...even on the earth, the angle is slightly less due to air resistance..but in one example i did earlier this yr (cant even remember the qn) the optimal angle worked out to be 38 degrees..i dont think that was 3u tho...maybe 4u?

 

[nike33]

in the example you gave you would assume gravity is the only force then possibly assume it..if you have time show range = ......and it follows sin2@=pi/2 is a max...@ = pi/4

 

[CrashOveride]

Originally posted by nike33
the optimal angle in EVERY case is not 45%...even on the earth, the angle is slightly less due to air resistance..but in one example i did earlier this yr (cant even remember the qn) the optimal angle worked out to be 38 degrees..i dont think that was 3u tho...maybe 4u?

Maybe that was part of the resisted motion chapter in 4unit. Coz for 3u projectiles we assume no air resistance and gravity to be the sole force acting on the object.

 

[Xayma]

I also think situations like a change in height (eg throwing from a ledge/downhill) results in a different optimum angle.

 

[George W. Bush]

but wait, remember to take into account the curvature of the Earth

 

[Xayma]

Originally posted by George W. Bush
but wait, remember to take into account the curvature of the Earth

And the change in gravity as the object gets higher.

And any lift created by the object (unless it is a perfect sphere and hence wont have an angle of attack)

 

[Grey Council]

ha!

You have no idea!

I chucked a javelin 30m and came 5th in the opens.
Moral? I know what I'm talking about.

its all in the height at which you chuck it. My comrades assumed that optimal angle of projection was 45 degrees.

However, they are throwing the javelin about 1.6 metres from the ground.
So optimal angle of projection is NOT 45.

blah, this is stoopid, i think I should stop.

Originally posted by Xayma
I also think situations like a change in height (eg throwing from a ledge/downhill) results in a different optimum angle.

Methinks thats true. In fact, I'm quite sure thats true.

 

[CrashOveride]

How do we work out the opt. angle of projection in those cases?