Projectile Motion (1 Viewer)

[Skeptyks]

Hey,
Just wondering if anyone had any tips for recalling the formulas for this topic.. There are so many formulas and their nothing like the parametric ones :/

Thanks.

 

[AbsoluteValue]

Hey,
Just wondering if anyone had any tips for recalling the formulas for this topic.. There are so many formulas and their nothing like the parametric ones :/

Thanks.

What do you mean by memorising formulae, in maths you don't memorise shit, you derive it.

 

[Skeptyks]

What do you mean by memorising formulae, in maths you don't memorise shit, you derive it.

That is true but some questions don't require you to derive the formula and rather just spit it out...

EDIT: Which, I guess, would make it much easier and faster when answering a particular question like that.

 

[Timske]

Vertical and Horizontal motion
Equation of trajectory
Range
Max Height
Max Range

 

[Sanjeet]

That is true but some questions don't require you to derive the formula and rather just spit it out...

EDIT: Which, I guess, would make it much easier and faster when answering a particular question like that.

Which questions require you to spit it out? I highly doubt that...

 

[Skeptyks]

Which questions require you to spit it out? I highly doubt that...

Sorry, for example, in 4U: A particle is projected under gravity from a point O. Prove that the horizontal distance between the two points on the path of the particle, which are at a vertical distance above O equal to 1/n of the greatest height, H, is R[(1-(1/n))^1/2] where R is the horizontal range. BTW sorry for the slow response to my own question but I want to finish a past paper tonight and so I won't be typing the working out completely.
We would 'spit out' H = v^2sin^2(alpha)/2g which would mean h = v^2sin^2(alpha)/2ng as h = (1/n)H. When letting y = h, we would get t(1) and t(2) which means v^2sin^2(alpha)/2ng = Vsin(alpha)t -1/2gt^2 and so on... TL;DR, it would be quicker to know the formula for the max height reached as opposed to deriving it from the max range formula.

Thanks in advance :]

 

[nightweaver066]

Sorry, for example, in 4U: A particle is projected under gravity from a point O. Prove that the horizontal distance between the two points on the path of the particle, which are at a vertical distance above O equal to 1/n of the greatest height, H, is R[(1-(1/n))^1/2] where R is the horizontal range. BTW sorry for the slow response to my own question but I want to finish a past paper tonight and so I won't be typing the working out completely.
We would 'spit out' H = v^2sin^2(alpha)/2g which would mean h = v^2sin^2(alpha)/2ng as h = (1/n)H. When letting y = h, we would get t(1) and t(2) which means v^2sin^2(alpha)/2ng = Vsin(alpha)t -1/2gt^2 and so on... TL;DR, it would be quicker to know the formula for the max height reached as opposed to deriving it from the max range formula.

Thanks in advance :]

In tests, they either ask you to derive it, or they'll supply the equations.

In textbooks, they don't specify, so do whichever you like (don't derive if you're already confident and feel that you'll just be wasting time).

 

[Skeptyks]

In tests, they either ask you to derive it, or they'll supply the equations.

In textbooks, they don't specify, so do whichever you like (don't derive if you're already confident and feel that you'll just be wasting time).

ahh, to be honest I havn't looked at any 4U past questions on mechanics so I didn't know that they would supply the equations/ask you to derive it specifically. Thanks :]