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Very basic projectile motion help (1 Viewer)

EkGew

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Due to our class having a student teacher for this topic, i didn't actually learn it (learning it now)

Problem with this question -
A coastal defence cannon fires a shell horizontally from the top of a 50.0 m high cliff, directed out to sea, with a velocity of 60.0 m s^-1. Calculate the range of the shell's trajectory.
 

shimmerz_777

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things to note.

1.start by dividing the problem into two seperate motions, one verticle and one horizontal i denote horizontal as x and verticle as y, and angle will = @

2. using triganometry, you can find the seperate motions of these two things. construct a right angle triangle with the angle of inclination, and the original velocity, V as the hypotenuse. your initial y velocity is, v= Vsin@, and your initial x velocity is Vcos@

3. you should know that the x motion does not undergo any acceleration, but the y component is subjected to gravity, so in its case, a = -9.8m/s^2

4. from this, you can discover everything you need to know, provided you have enough info through your standard equasions for acceleration, displacement and everything i.e v = u + at .....

5. you will have to combine results from your x and y components, such as time together to find your answers (usually)

in this question, @ = 0, so initially, y_v = 0, and x_v = 60

first find how long it takes the shell to hit the ground, knowing that a = -9.8, and u=0 and that it needs to travel 50 m

using s= ut + 1/2 x at^2
-50 = 0t + 1/2 x -9.8t^2
-100/-9.8 = t^2
t= 3.1943....

now since we have the time of flight to find the range you multiply this by the horizontal speed ( which remains constant), as d=vt

60 x 3.1943... = 191.66m
or if u want the answer to nearest meter 192m's

make sense?
 
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del pietro

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personally i find it so much easier and quicker doing projectile motion questions using the calculus method, which we learn in 3 unit maths..
 

del pietro

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would the markers acknowledge this form of working out in the marking??

and yeh the answer is 191.66m
 

Riviet

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del pietro said:
would the markers acknowledge this form of working out in the marking??
It's generally frowned upon if you use 3u methods learnt from the maths extension course, you should just stick to the formulae in physics.
 

helper

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Rubbish Riviet, any correct method is accepted.
That is the maths method, the formulas in the HSC data sheet, traditional formulas using suvat or the rote learnt formulas based on the assumption of flat ground if appropriate.

That is why the 4 mark question in 2004 was a joke. A student who wrote learn't the range formula gained full marks for 3 lines, even though commonly students who learn this are students that don't understand components, while a student who understands components and the physics makes a simple mistake and didn't gain full marks.
 

rama_v

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http://community.boredofstudies.org/18/physics/84365/projectile-motion-calculations.html?highlight=3+unit

香港! said:
I sent an email asking if I would lose marks if I used 3 Unit Mathematics techniques to answer projectile motion and here's the reply:

"The general instruction given on the HSC Physics paper for answers
involving calculations is that all relevant working should be shown.
All legible responses following this instruction to obtain the correct
answer will be awarded full marks. While the HSC Physics course does not
require the use of calculus, responses that use it correctly will not lose
marks. Responses showing appropriate working, but an incorrect answer
gained through faulty arithmetic may still score some marks.
Please do not hesitate to contact me if you have any further enquiries.

Margaret Baldry
Senior Assessment Officer, Science
"
Just wanted to make it clear to everyone now:)
 

Sober

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del pietro said:
alright thats good.. thanx for that rama v..
If you use calculus and make an error there is a much greater chance you will lose marks than if you had used the simple formulas provided and made an error.

If you are smart enough to understand 3 unit maths then why must you find these simple formulas hard? They are essentially the same, just sidestepping the calculus.
 

rama_v

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Sober said:
If you use calculus and make an error there is a much greater chance you will lose marks than if you had used the simple formulas provided and made an error.

If you are smart enough to understand 3 unit maths then why must you find these simple formulas hard? They are essentially the same, just sidestepping the calculus.
That is true. But if people want to do it that way, and it assists understanding, then why not
 

gamecw

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Sober said:
If you are smart enough to understand 3 unit maths then why must you find these simple formulas hard? They are essentially the same, just sidestepping the calculus.
3unit's way is mucher quicker. And those formulea are not hard, it just takes more steps to solve the same question..

Eg.
Q. 9+9+9+9+9+9+9=?
If u know u can simply obtain the answer by doing 9x7=63, why go through all the trouble adding the 9s.
 

JBakaka

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even though i agree that you prob can use the 3u maths formulas, i wouldnt just to stick to the safe side because you can get physics markers who don't know that method and could mark you down
for me, i just stick to physics formulae in physics and maths formulae in maths otherwise i get confused
 

I M J O S H

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Ok so Ive got an extended version of that question....

Q. To increase the range of the shell in the previous question the cannon is lifted, so that it now points up at an angle of 45 degrees.
Find the new range.

So the cannon is up on a 50m cliff, and fires off a ball at 60ms at an angle of 45 degrees.

I cant do this question for some reason.:burn:
 

ianc

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It's just about getting the right ux and uy

Have a look at this:


Then you just do it like any other question. If you still need help, don't hesitate to ask.
 
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I M J O S H

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I didnt word it very well, my bad.

I know how to resolve the vectors and all that jazz, just whats throwing me is the fact that its up on a 50 m cliff and then firing a ball at 60ms at an angle of 45 degrees. Like, I can do it if its on a flat surface but Im unsure of what to do with the 50m...

Any help would be great, cheers.
 

ianc

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You just add 50 to the displacement on the y axis

so sy = uyt + (1/2)ayt2 + 50
(note: ay = -9.8 ms-2)

to get the range, you want to find the time when sy = 0

Then you substitute it into sx = ux t


ALTERNATIVELY:

you could just have so sy = uyt + (1/2)ayt2

and then find the time when sy = -50 (because it would have fallen 50 metres)



Hope this helps!
 

I M J O S H

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Rofl, I swear thats what Ive been doing, yet I continue to get the wrong answer. :burn:

In all your wisdom might you post a worked answer? It would be greatly appreciated.
 

ianc

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sy = uyt + (1/2)(-9.8)t2 + 50
0 = - 4.9t2 + (60/√2)t + 50
- 4.9t2 + (30√2)t + 50 = 0

We need to use the quadratic formula,
so t = [ -30√2 ± √(1800 + 980) ] / [-9.8]
since t>0, t = [ -30√2 - √(1800 + 980) ] / [-9.8]

work this out and store the value in your calculator

then substitute it into

sx = uxt
= (30√2)t

and i worked it out to be 412m (nearest metre)

...hope this is the right answer
 

I M J O S H

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Cheers.

And yes you got the right answer.

This is really gonna piss you off, but you reckon you could do it using the projectile motion equations instead of the quadratic :)

If you dont want to, completely understand, thanks for the help you gave :)
 

JBakaka

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i think that to use the proj. motion equations (and not the quad. formula), youd have to get v (final velocity), which is not possible to get without calculating the time of flight 't' or going on some crazily unnecessarily complicated route
 

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