Harder Projectile Motion Q (1 Viewer)

[Makematics]

Hey guys, so there is this question in cambridge 4U, and i just can't get it. It would be much appreciated if you could help me out ive looked at the solutions that are available online and they are shit and dont make any sense, so yeah...

A particle is projected with speed V from the top of a cliff of height h above sea level. Find the greatest horizontal distance the particle can cover before landing in the sea.

 

[braintic]

Hey guys, so there is this question in cambridge 4U, and i just can't get it. It would be much appreciated if you could help me out ive looked at the solutions that are available online and they are shit and dont make any sense, so yeah...

A particle is projected with speed V from the top of a cliff of height h above sea level. Find the greatest horizontal distance the particle can cover before landing in the sea.

Find the Cartesian equation of the projectile's path, let y=0, and use the quadratic formula to write x as a function of theta (taking only the +ve case). Then use calculus to maximise x.

 

[Makematics]

i tried that. got me nowhere. even after 4 pages of working i couldnt get anywhere. it's probably a matter of forcing out the algebra

 

[Makematics]

BTW if anyone wants some help, the answer is (V/g) sqrt (V^2 +2gh)

 

[Makematics]

ok well ive gotten down to cos (2alpha)=gh/(V^2 +gh) idk if that helps though :/

 

[HeroicPandas]

can u say that DELTA = 0? because there is only 1 solution for x

 

[Makematics]

can u say that DELTA = 0? because there is only 1 solution for x

as in the discriminant? i dont think so, since the other solution is negative and exists, it's just that we dont want it. i think...

 

[RealiseNothing]

as in the discriminant? i dont think so, since the other solution is negative and exists, it's just that we dont want it. i think...

This.

 

[Makematics]

the most amazing thing just happened, i did it!!! for some reason before i put this post up, i was really close to getting it, but i just stopped, because it was getting too ugly. moral of the story is never stop in the middle of a proof! i would write up a solution on latex for anyone who is curious, but i would be here till tomorrow. also if anyone finds a quicker way than the method which braintic described, let me know. because that was just brutal algebra and trigonometry.

 

[Menomaths]

Lmao attempted to solve this question thinking it was Physics...but then I realized...

 

[Sy123]

After quadratic formula and that business:



Taking the natural logarithm of both sides then differntiating gives us an easier time, also notice that when we do so the second bracket becomes a standard integral:



letting R'(a) = 0, then simplifying, making use of pythagorean trig identity, we arrive after a long haul:



Now setting up a right angled triangle, we find that:



Substituting this into R(a), then fiddling around, noticing that the stuff in the square root miraculously forms a perfect square:



Putting all of this crap in we get a cancellation, we arrive at the answer.

 

[Makematics]

ok so in other words, there is no way around the painfully long algebra... but that's a nice method you have there. i used the same method that braintic suggested and it worked out too!

 

[Makematics]

the sad thing is it looks like a reasonable 3U question at first glance...

 

[Sy123]

ok so in other words, there is no way around the painfully long algebra... but that's a nice method you have there. i used the same method that braintic suggested and it worked out too!

I am wondering whether there is a differential equation type solution. If:



Then we can see that:



So that means, if we can prove the above differential equation perhaps by thinking physically, using forces and whatnot then it may be a possible solution.

I'm just throwing ideas out there lol it may not have any physical relevance whatsoever.

 

[rural juror]

actually you don't need a lot of algebra if you do it the simple way,

right down your equation of motion

tanOx - x^2sec^Og/2v^2 + h = 0 and then implicitly differentiate that. You get the condition that tanO = V^2/gx, which if you sub that in, kills it. its much easier than fiddling around with square root signs

 

[Sy123]

actually you don't need a lot of algebra if you do it the simple way,

right down your equation of motion

tanOx - x^2sec^Og/2v^2 + h = 0 and then implicitly differentiate that. You get the condition that tanO = V^2/gx, which if you sub that in, kills it. its much easier than fiddling around with square root signs

That makes much more sense, nice.

 

[Makematics]

actually you don't need a lot of algebra if you do it the simple way,

right down your equation of motion

tanOx - x^2sec^Og/2v^2 + h = 0 and then implicitly differentiate that. You get the condition that tanO = V^2/gx, which if you sub that in, kills it. its much easier than fiddling around with square root signs

wow nice, repped.

 

[Makematics]

what i did was a good algebraic challenge nonetheless