Advanced Mechanics (1 Viewer)

[ExtremelyBoredUser]

Hey guys, so I've been doing the first part of the module and projectile motion questions. I've encountered a question (Pearson Review 2.2 Q4) which asks to find the initial velocity given the Horizontal Range and the Angle of projection. Am I allowed to use the Horizontal Range equation?

R = v^2sin2a/g

I originally did the question using this formula and I got the correct answer however I am wondering if you are allowed to use that equation? If it is not, how would you attempt the question using basic suvat equations. Cheers.

Question for reference;

 

[A1La5]

I know you aren't allowed to use Physics formulas in Maths, but I'm not sure about whether you can use Maths formulas in Physics. Just to be on the safe side I would stay away from it and just use the basic SUVAT equations, but I think you can. The questions in the HSC are catered towards using those equations (as the examiners cannot assume that everyone who does Physics does Extension 2 Mathematics).

What I would do is to decompose the initial velocity (call it "U") into vertical and horizontal components using the given launch angle. Then assuming the parabolic trajectory is symmetrical about its axis of symmetry I would note that the time taken to reach maximum height is half it's time of flight. Furthermore I would also note that the projectile's vertical velocity at maximum height is zero. You can use this information to find an expression for the time taken to reach maximum height in terms of the projectile's vertical component and gravity (9.8m/s/s) using the first SUVAT equation.

Once the projectile touches base with the ground again it reaches its time of flight. Using the given range and the formula s = Ucos(angle) x time of flight (or s = Ucos(angle) x 2 x time to reach maximum height), you can substitute and solve for U.

Finding the value of U would be a matter of using the Pythagorean Theorem once you have the horizontal and vertical components Nope, you're meant to use simultaneous equations..

EDIT: Just did the question, and I managed to get to the right answer using the SUVAT equations.

 

[jimmysmith560]

Hey, I'm not sure if this will be of help, but I thought I'd include working out to this question:



I hope this helps!

 

[ExtremelyBoredUser]

I know you aren't allowed to use Physics formulas in Maths, but I'm not sure about whether you can use Maths formulas in Physics. Just to be on the safe side I would stay away from it and just use the basic SUVAT equations, but I think you can. The questions in the HSC are catered towards using those equations (as the examiners cannot assume that everyone who does Physics does Extension 2 Mathematics).

What I would do is to decompose the initial velocity (call it "U") into vertical and horizontal components using the given launch angle. Then assuming the parabolic trajectory is symmetrical about its axis of symmetry I would note that the time taken to reach maximum height is half it's time of flight. Furthermore I would also note that the projectile's vertical velocity at maximum height is zero. You can use this information to find an expression for the time taken to reach maximum height in terms of the projectile's vertical component and gravity (9.8m/s/s) using the first SUVAT equation.

Once the projectile touches base with the ground again it reaches its time of flight. Using the given range and the formula s = Ucos(angle) x time of flight (or s = Ucos(angle) x 2 x time to reach maximum height), you can substitute and solve for U.

Finding the value of U would be a matter of using the Pythagorean Theorem once you have the horizontal and vertical components Nope, you're meant to use simultaneous equations..

EDIT: Just did the question, and I managed to get to the right answer using the SUVAT equations.

Hey sorry for the late reply. My mistake was in the assumptions as I am more used to Math-style projectile in which acceleration is constantly changing and there is a resistive force so I did not assume the symmetry that would've given the answer to the question through substitution of equations. I did the same thing as you up to your third paragraph but its pretty simple now looking as you would just substitute time of flight equation and vertical component into one of the displacement ones and solve for u.

 

[ExtremelyBoredUser]

/

Hey, I'm not sure if this will be of help, but I thought I'd include working out to this question:

View attachment 32104

I hope this helps!

Yeah that's what I originally did but I realised that would've been out of the HSC physics syllabus. I'm unsure if you can do that or not because personally I prefer that equation since it makes computing way more faster but it is what is ig

 

[jazz519]

Hey guys, so I've been doing the first part of the module and projectile motion questions. I've encountered a question (Pearson Review 2.2 Q4) which asks to find the initial velocity given the Horizontal Range and the Angle of projection. Am I allowed to use the Horizontal Range equation?

R = v^2sin2a/g

I originally did the question using this formula and I got the correct answer however I am wondering if you are allowed to use that equation? If it is not, how would you attempt the question using basic suvat equations. Cheers.

Question for reference;
View attachment 32098

The problem with using these other equations is that you run the risk of if you get it wrong you might not get many marks, because there is not much working associated with them as they are already derived relationships. All the questions in HSC Physics can be solved using only the formulas taught so I would stick to these to be on the safe side.

 

[ExtremelyBoredUser]

The problem with using these other equations is that you run the risk of if you get it wrong you might not get many marks, because there is not much working associated with them as they are already derived relationships. All the questions in HSC Physics can be solved using only the formulas taught so I would stick to these to be on the safe side.

Yeah I figured, but on the off chance, if you were to use the equation and get it correct, would you still receive full marks or would you be penalised? Not implying that I'll do it since I want to get used to suvat equations again but out of curiosity.

 

[jazz519]

Yeah I figured, but on the off chance, if you were to use the equation and get it correct, would you still receive full marks or would you be penalised? Not implying that I'll do it since I want to get used to suvat equations again but out of curiosity.

If you derive it then yeah you would probably get the marks, but like lets say for example it's a 4 mark question and you just quote an equation without deriving it and your working is like 2 lines, you probably won't get the full marks. It's similar to how in maths the working is important to show

 

[ExtremelyBoredUser]

If you derive it then yeah you would probably get the marks, but like lets say for example it's a 4 mark question and you just quote an equation without deriving it and your working is like 2 lines, you probably won't get the full marks. It's similar to how in maths the working is important to show

Ah okay that makes sense, thanks for clearing that up.