Projectile Motion - Weird substitution for Time of Flight

[Jefferson]

Hi all,

For the following question, I partly substituted time,t,  (in attachment below) instead of doing it normally like how you would to find the initial velocity given angle, range and  final height using the cartessian equation.

This way seems to work quite nicely since the 'u' cancels, and to find the initial speed (u) with the answer for time (t) is also quite quick afterwards.

Are there any issues with taking this approach in the HSC?
Does it have any limitations/assumptions or will it work for all similar cases?

[RuiAce]

You're basically just trying to pull off a clever trick for the sake of doing later algebra later on right?

Yeah, that's perfectly acceptable. If there was another part that explicitly wanted say \(y\) in terms of \(x\) then you would be penalised. But here, because it isn't mentioned, provided your computations are not incorrect being clever with algebra is not a problem.

Just be careful when handling square roots, because your scenario is more or less expecting the cannonball to hit the wall during its descent. In theory, of course the cannonball could hit the wall during the ascent instead. (I presume it wouldn't be able to though in this situation.)

[Jefferson]

You're basically just trying to pull off a clever trick for the sake of doing later algebra later on right?

Yeah, that's perfectly acceptable. If there was another part that explicitly wanted say \(y\) in terms of \(x\) then you would be penalised. But here, because it isn't mentioned, provided your computations are not incorrect being clever with algebra is not a problem.

Just be careful when handling square roots, because your scenario is more or less expecting the cannonball to hit the wall during its descent. In theory, of course the cannonball could hit the wall during the ascent instead. (I presume it wouldn't be able to though in this situation.)

Hi, RuiAce.

Thanks for clarifying, though why would this scenario assumes that the cannonball hits the castle during descent?

With the angle being locked in, I used the range (x value) and height (y value) of the castle simultaneously to work out the time and initial speed of the cannonball. At this point, I don't know (omitting the diagram) whether it'll collide with the castle on the way up or down.

Therefore, I think it should also work even if it collides with the castle during its ascent. (only 1 y value for every x value, so 1 answer for time and inital velocity).

I'd imagine that to figure out whether it collided with the castle during the ascent or descent, you'd have to find the time when it is at maximum height and compare it to the time when it collides (less than t-max = ascent, more than t-max = descent).

Or otherwise, perhaps easier, is to find Vy^. (vertical velocity at time of collision) to see if it's positive (still going up, i.e. ascent) or negative (going down, i.e. descent).

Thanks again.

[RuiAce]

Hi, RuiAce.

Thanks for clarifying, though why would this scenario assumes that the cannonball hits the castle during descent?

With the angle being locked in, I used the range (x value) and height (y value) of the castle simultaneously to work out the time and initial speed of the cannonball. At this point, I don't know (omitting the diagram) whether it'll collide with the castle on the way up or down.

Therefore, I think it should also work even if it collides with the castle during its ascent. (only 1 y value for every x value, so 1 answer for time and inital velocity).

I'd imagine that to figure out whether it collided with the castle during the ascent or descent, you'd have to find the time when it is at maximum height and compare it to the time when it collides (less than t-max = ascent, more than t-max = descent).

Or otherwise, perhaps easier, is to find Vy^. (vertical velocity at time of collision) to see if it's positive (still going up, i.e. ascent) or negative (going down, i.e. descent).

Thanks again.

To be fair it's harder to figure it out by inspection. (More or less because one thing you've specified is important - the angle is locked in.)

Intuition tells me that usually if there's an ascent and descent possibility, the relevant quadratic you need to solve should yield two distinct positive solutions. It further tells me that if one thing ends up being negative, usually the ascent possibility is ruled out. (Usually the ascent-possibility corresponds to the negative square root when doing any quadratic formula stuff, and the positive square root corresponds to the descent.)

But of course, this kind of intuition could easily be wrong, and I'd be happy to accept otherwise if you did find some counterexample. That's what I'd gather though, personally.

[Jefferson]

To be fair it's harder to figure it out by inspection. (More or less because one thing you've specified is important - the angle is locked in.)

Intuition tells me that usually if there's an ascent and descent possibility, the relevant quadratic you need to solve should yield two distinct positive solutions. It further tells me that if one thing ends up being negative, usually the ascent possibility is ruled out. (Usually the ascent-possibility corresponds to the negative square root when doing any quadratic formula stuff, and the positive square root corresponds to the descent.)

But of course, this kind of intuition could easily be wrong, and I'd be happy to accept otherwise if you did find some counterexample. That's what I'd gather though, personally.

Hi RuiAce

Sorry it took a while, I couldn't find the right question so I made up my own, hence the weird/convenient numbers. (See Attachment)
I made sure that there are no contradictions in any value (unlike most textbooks ), so no need to worry on that end.

The question has exactly the same amount of information as the one in this post, with the key difference being where the ball hits the castle on its path.

Since there is only one y value for each x value, there can only be one possible time for this occurence. This means that a quadratic will never form to give more than one positive value, as there can only be one time and one x value at this point. (hence always a plus and minus if it's a quadratic).
I don't quite understand what you said about the negative answer in the quadratic ruling out the ascending case, as I have not yet seen anything logical that may suggest this to be the case.

The converse is obviously not true though. One y value can take on (up to) 2 x values due to the symmetrical shape of the parabola.
A quadratic would form when you solve for the height, y, being 20m, which you will get 43.017m for x at 0.993s (ascend) and 177.908m at 4.019 seconds (descend).


The fact that you know where the castle is (x) helps you find out whether the ball will collide with it during its ascension or descension, from my observation.
Thanks for reading, again.