Mod 5 physics (3 Viewers)

[Tryingtodowell]

Ok now since Im done with projectile motion in mod 5 I havent come across this formula:


Is this from the old syllabus or something that pops in mod 5 later on? idk I searched the syllabus/ textbooks and couldnt find this so just making sure

thanks

 

[Average Boreduser]

Ok now since Im done with projectile motion in mod 5 I havent come across this formula:
View attachment 44457

Is this from the old syllabus or something that pops in mod 5 later on? idk I searched the syllabus/ textbooks and couldnt find this so just making sure

thanks

its a formula u can derive using integration. i think u can also derive it using the suvat eqns asw but im unsure. that there is the formula for the maximum height approached by the proj motion.

 

[iloveeggs]

its like a handy tool ig but you're better off not memorising it imo. maybe for practise you could derive it so you can see where it comes from

 

[Tryingtodowell]

its like a handy tool ig but you're better off not memorising it imo. maybe for practise you could derive it so you can see where it comes from

is there a reason why not to memorise it? it doesnt seem that bad

 

[zzz5428]

You can't quote formulas not on the data sheet -> must prove them

 

[Tryingtodowell]

ohk

 

[Tryingtodowell]

You can't quote formulas not on the data sheet -> must prove them

Im guessing you have done hsc physics
have you used this formula a lot during ur physics journey? want to see how important it is

 

[iloveeggs]

You can't quote formulas not on the data sheet -> must prove them

yeah this is the reason. you can't use it straight up in the hsc. its only useful for multiple choice bc no working is needed (unlikely tho) and maybe its useful for quickly checking your answer for a longer question ig

 

[zzz5428]

Not particularly, projectile motion questions tend to be test only once or twice throughout the entire paper and are usually considered 'lower band' questions

 

[moonsuyoung]

Ok now since Im done with projectile motion in mod 5 I havent come across this formula:
View attachment 44457

Is this from the old syllabus or something that pops in mod 5 later on? idk I searched the syllabus/ textbooks and couldnt find this so just making sure

thanks

It's used for symmetrical projectiles only I'm pretty sure. Like people have said above, any equations in physics that are not on formula sheet you need to remember to derive or you lose marks. But just remember for multiple choice imo.

 

[C_master]

It's used for symmetrical projectiles only I'm pretty sure. Like people have said above, any equations in physics that are not on formula sheet you need to remember to derive or you lose marks. But just remember for multiple choice imo.

oi you done w mod 5?

 

[retroojh]

i think its to find maximum height for the projectile,

since vy²= uy² + 2gh (from v² = u² + 2as, subbed in acceleration as gravity and displacement as height)
vy² = 0 since you reach max height when vy = 0
so, 0 = uy² + 2gh
rearrange for h
h = -uy² / 2g
(im taking upwards as positive so the negatives cancel out)

h = uy² / 2g
and since uy² = usinθ (resolving initial velocity)

h = (usinθ) ² / 2g
they just expanded the bracket lol
this is for the case when it lands from ground to ground (launch height = landing height)
but if it has a launch height, make sure to add it

so itll be h = (usinθ) ² / 2g + launch height

 

[liamkk112]

Ok now since Im done with projectile motion in mod 5 I havent come across this formula:
View attachment 44457

Is this from the old syllabus or something that pops in mod 5 later on? idk I searched the syllabus/ textbooks and couldnt find this so just making sure

thanks

you don't need this in physics hsc
if you're launching the projectile at altitude 0 on a flat surface, then the max height occurs at time of flight /2 by symmetry, which is all that's assessed
if ever they ask for something else, there'll be something else that indicates when the maximum will occur, but you definetly don't need to know that formula (which is from calculus based projectile motion, tested in 3u, but even then you have to derive it when you use it usually)

 

[nonya2000]

Ok now since Im done with projectile motion in mod 5 I havent come across this formula:
View attachment 44457

Is this from the old syllabus or something that pops in mod 5 later on? idk I searched the syllabus/ textbooks and couldnt find this so just making sure

thanks

No. You don't even need to know this formula. If a question requires it they will give it anyway.

 

[StudyNotesTips]

The formula you're referencing (often represented as R=v2sin⁡2θgR = \frac{v^2 \sin 2\theta}{g}R=gv2sin2θ for the range of a projectile) is indeed associated with projectile motion, but it may not be explicitly stated in the current HSC Physics syllabus or in the textbooks you’ve reviewed for Module 5.

Here's a bit of context:

1. Old Syllabus: This formula has been commonly used in earlier iterations of physics courses, including the old HSC syllabus. It gives the horizontal range of a projectile launched at an angle θ\thetaθ with an initial velocity vvv.
2. Current Syllabus: While the concepts of projectile motion are covered in Module 5, the specific formulas and their applications might not be emphasized in the same way. Instead, you're expected to derive such formulas based on the principles of kinematics and dynamics.

Why You Might Not See It:

• The current focus may be more on understanding the underlying principles rather than memorizing specific formulas.
• Your syllabus might prioritize concepts over specific mathematical expressions.

What to Do:

• If you're comfortable with the derivation of projectile motion and understand how to calculate range, height, and time of flight, you're likely on the right track!
• If you think you might encounter this formula in later topics or assessments, it could still be helpful to familiarize yourself with it and its derivation.

 

[nonya2000]

The formula you're referencing (often represented as R=v2sin⁡2θgR = \frac{v^2 \sin 2\theta}{g}R=gv2sin2θ for the range of a projectile) is indeed associated with projectile motion, but it may not be explicitly stated in the current HSC Physics syllabus or in the textbooks you’ve reviewed for Module 5.

Here's a bit of context:

1. Old Syllabus: This formula has been commonly used in earlier iterations of physics courses, including the old HSC syllabus. It gives the horizontal range of a projectile launched at an angle θ\thetaθ with an initial velocity vvv.
2. Current Syllabus: While the concepts of projectile motion are covered in Module 5, the specific formulas and their applications might not be emphasized in the same way. Instead, you're expected to derive such formulas based on the principles of kinematics and dynamics.

Why You Might Not See It:

• The current focus may be more on understanding the underlying principles rather than memorizing specific formulas.
• Your syllabus might prioritize concepts over specific mathematical expressions.

What to Do:

• If you're comfortable with the derivation of projectile motion and understand how to calculate range, height, and time of flight, you're likely on the right track!
• If you think you might encounter this formula in later topics or assessments, it could still be helpful to familiarize yourself with it and its derivation.

how gay are you?

 

[nonya2000]

think you meant to type

 

[StudyNotesTips]

how gay are you?

You could respond with humor or lightheartedness, depending on your relationship with the person. Here’s a playful response:

"Well, I’m not in the rainbow parade, but I’m definitely here to support and spread good vibes! How about you?"

This keeps it fun while engaging the other person in conversation.

 

[C_master]

You could respond with humor or lightheartedness, depending on your relationship with the person. Here’s a playful response:

"Well, I’m not in the rainbow parade, but I’m definitely here to support and spread good vibes! How about you?"

This keeps it fun while engaging the other person in conversation.

 

[wizzkids]

It is another form of the relation v²=u²+2as.
It is only considering the vertical component of velocity v_y = v sin θ
You should be able to derive this relation easily.