Physics Questions Help Please (1 Viewer)

[Bisu]

A stone of mass 0.2 kg was thrown horizontally at 20m/s from the top of a cliff of 19.6m. It strikes the ground at a certain istance from the foot of the cliff.

a) Calculate the distance between the foot of cliff and the point of an impact of the stone with the ground:

b) Calculate the vertical velocity the stone hits the ground

c) calculate the velocity the stone hits the ground.

Worknig too please..

I just can't get my head around the horizontal part..Only used to doing questions with an angle given so help all appreciated

 

[natalie116]

do you have the answers?
i'm doing the question but don't want to give you the wrong working.

 

[Schoey93]

A stone of mass 0.2 kg was thrown horizontally at 20m/s from the top of a cliff of 19.6m. It strikes the ground at a certain istance from the foot of the cliff.

a) Calculate the distance between the foot of cliff and the point of an impact of the stone with the ground:

b) Calculate the vertical velocity the stone hits the ground

c) calculate the velocity the stone hits the ground.

Worknig too please..

I just can't get my head around the horizontal part..Only used to doing questions with an angle given so help all appreciated

Ooh! I was reading about this just yesterday! Let me go get that book, Mathematics: A Very Short Introduction.

BTW I'm not taking the piss, that's a real book, give me a minute or two.

 

[Schoey93]

This is directly from the book I mentioned:

"How to throw a stone

"The best compromise, which can be worked out using a combination of Newtonian physics and some elementary CALCULUS, turns out to be as neat as one could hope for under the circumstances: the direction of the stone as it leaves your hand should be upwards at an angle of 45 degrees to the horizontal.The same calculations show that the stone will trace out a PARABOLIC curve as it flies through the air, and they tell you how fast it will be travelling at any given moment after it leaves your hand."

Clearly, Natalie, that is technically an HSC MATHEMATICS 2U problem, rather than a (simpler) HSC Physics problem. I can just say this: I hope you're doing 2U Maths, because if this problem is typical of what you are learning in HSC Physics, you're screwed if you don't know how to do calculus... and if you don't know how to do it, you'd better start learning, either from Excel Mathematics Preliminary by Lyn Baker or (less preferable) Maths in Focus by Margaret Grove.


Hope this helps...please notice that the ANGLE IS 45 DEGREES!

 

[darkchild69]

A stone of mass 0.2 kg was thrown horizontally at 20m/s from the top of a cliff of 19.6m. It strikes the ground at a certain istance from the foot of the cliff.

a) Calculate the distance between the foot of cliff and the point of an impact of the stone with the ground:

b) Calculate the vertical velocity the stone hits the ground

c) calculate the velocity the stone hits the ground.

Worknig too please..

I just can't get my head around the horizontal part..Only used to doing questions with an angle given so help all appreciated

The angle is given

"A stone of mass 0.2 kg was thrown horizontally"

The angle is 0

So that means the initial velocity (20ms-1) is the initial horizontal velocity (Ux) and the initial vertical velocity (Uy) = 0

I'm sure you can do it now

for (a) just use s = ut + 1/2at^2

where:
s = -19.6
u = 0
a = -9.8

then solve for t (which will be sqrt 4 = 2s)

then use Sx = Ux*t
Sx = 20*2
Sx = 40m

(b)

v = u + at
v = 0 + -9.8*2
v = -19.6ms-1

(c)

Vy = -19.6ms-1
Vx = Ux = 20ms-1

You can do this one, i can't do it without a calculator, lol...

Just remember, as it is velocity it also needs the angle with which it strikes the ground (seeing as Vy and Vx are nearly equal, you should expect the angle to be around 45 degrees)

 

[Schoey93]

The angle is given

"A stone of mass 0.2 kg was thrown horizontally"

The angle is 0

So that means the initial velocity (20ms-1) is the initial horizontal velocity (Ux) and the initial vertical velocity (Uy) = 0

I'm sure you can do it now

for (a) just use s = ut + 1/2at^2

where:
s = -19.6
u = 0
a = -9.8

then solve for t (which will be sqrt 4 = 2s)

then use Sx = Ux*t
Sx = 20*2
Sx = 40m

(b)

v = u + at
v = 0 + -9.8*2
v = -19.6ms-1

(c)

Vy = -19.6ms-1
Vx = Ux = 20ms-1

You can do this one, i can't do it without a calculator, lol...

Just remember, as it is velocity it also needs the angle with which it strikes the ground (seeing as Vy and Vx are nearly equal, you should expect the angle to be around 45 degrees)

Yes, the angle at which the stone is THROWN is of course, ZERO. But the initial angle of the parabolic curve is 45 DEGREES> BTW Way to go! You just bastardized a simple maths question by making it TOO simple with physics formulae.

Please derive all of the formulae you mentioned to prove me wrong and to prove that you didn't bastardize the question. Can't do it, can you?

 

[natalie116]

Yes, the angle at which the stone is THROWN is of course, ZERO. But the initial angle of the parabolic curve is 45 DEGREES> BTW Way to go! You just bastardized a simple maths question by making it TOO simple with physics formulae.

Please derive all of the formulae you mentioned to prove me wrong and to prove that you didn't bastardize the question. Can't do it, can you?

mate, forget about the 45 degrees thing.
it just a physics question, and can be done with a couple of formulas. That's it, don't get so worked up

to original poster:
DW!! it took me a while to get my head around projectile motion. After these questions it doesnt get much harder ^^

 

[darkchild69]

Yes, the angle at which the stone is THROWN is of course, ZERO. But the initial angle of the parabolic curve is 45 DEGREES> BTW Way to go! You just bastardized a simple maths question by making it TOO simple with physics formulae.

Please derive all of the formulae you mentioned to prove me wrong and to prove that you didn't bastardize the question. Can't do it, can you?

WTF are you on about?

Do you even have any idea what the question is actually asking??

LMFAO

Please remind me what section we are in again?? On yeah PHYSICS

Just saw that you are in year 10. That explains why you have no idea what the original poster is asking. Stick to sections in which you actually have an idea and please dont give "solutions" to questions if you have no idea, it just confuses people when a person who actually knows what they are on about comes in and answers the questions.

thanks

Mate, these equations are piss easy to derive, anyone who has done / is doing 3u Mathematics can show versions of the derived equations, but I dont think you are quite up to understanding it, so i shan't waste my time. Look through any Physics textbook under "projectile motion formulae" and there they are

Don't worry mate, you'll get there one day

 

[natalie116]

Yes, the angle at which the stone is THROWN is of course, ZERO. But the initial angle of the parabolic curve is 45 DEGREES> BTW Way to go! You just bastardized a simple maths question by making it TOO simple with physics formulae.

Please derive all of the formulae you mentioned to prove me wrong and to prove that you didn't bastardize the question. Can't do it, can you?

you're in yr 10?!?!? get outta here!! dont go on blabbering stuff from a book you don't understand. go do physics in year 11 and then you will realise how you just embarrassed yourself.

 

[xMaFF]

(o_____________o)???

Woahh man.


I don't want to do physics for my prelims next year anymoreee

Haha, I'm kidding.
It's good to know that there are some really cool people here on the forums to help out.

 

[k02033]

you're in yr 10?!?!? get outta here!! dont go on blabbering stuff from a book you don't understand. go do physics in year 11 and then you will realise how you just embarrassed yourself.

dont tease him.. even if he didnt gasp the question, i think he has a good altitude towards maths/physics, very rare especially for someone in year 10. And hes reading a maths book way ahead of his syllabus, i think that is great.

 

[darkchild69]

Mate, you should do it. Prelim is a bit airy fairy, more like expanding a little on the Stage 5 course, but the HSC course is real interesting.

dont tease him.. even if he didnt gasp the question, i think he has a good altitude towards maths/physics, very rare especially for someone in year 10. And hes reading a maths book way ahead of his syllabus, i think that is great.

His attitude towards his studies may be good, but his attitude towards others on the forums sucks ass. You don't come out here on these forums and blast others, especially when you have nfi wtf you are on about.

He may be reading the textbook, but i don't think he is actually understanding the concepts behind it. Initial angle of the parabolic curve in this question is 45 degrees, but yet the initial angle of the trajectory of the object undergoing parabolic motion is zero? wtf?

I could give a 4u textbook to my 10 year old nephew, just because he would read it, doesn't mean he could do anything with it.

 

[k02033]

Initial angle of the parabolic curve in this question is 45 degrees,

i dont get that. could you explain? who/how did they arrive at that conclusion? what parabolic function are we talking about?

I could give a 4u textbook to my 10 year old nephew, just because he would read it, doesn't mean he could do anything with it.

did you try it?

 

[darkchild69]

i dont get that. could you explain? who/how did they arrive at that conclusion? what parabolic function are we talking about?

Lol, that's my point exactly.. He came into this thread and began reciting words from a textbook example as if it applied to the question asked by the original poster, which it clearly did not. Then he began mouthing off at me as if his recitation still held some merit.

did you try it?

Yes

And he was better at Conics in 3 days than I ever was

But i owned him in resisted motion, fourier analysis, Gauss and Stokes theorems and the conical pendulum

 

[khorne]

Yes, the angle at which the stone is THROWN is of course, ZERO. But the initial angle of the parabolic curve is 45 DEGREES> BTW Way to go! You just bastardized a simple maths question by making it TOO simple with physics formulae.

Please derive all of the formulae you mentioned to prove me wrong and to prove that you didn't bastardize the question. Can't do it, can you?

Now, firstly, you'll notice I am in year 10, but please don't let any pre-conceptions mar your judgment =]
Wtf? The reference in the book was talking about an old optimisation problem, of how "best" to throw a stone, which, in perfect conditions, with the only force acting on it being gravity, would be at a 45 degree angle.

Of course, this is not talking about that.

These formulas (please note, ae is not required) were derived in school.

As for your point, these formulas are meant to make it easier, it's not bastardising (weren't not in America, now, are we?) the question, it's solving it...

As for the formula's used, I will prove them, for your benefit.

ICBS: a = v-u/t

therefore, v = u + at (1)

Av. Vel. = v-u/2 = s/t

therefore, s = (u+v/2) t

but t = v-u/a

so s = (u+v/2)(v-u/a)

s = v^2 - u^2/2a

therefore:

v^2 = u^2 + 2as

Finally, v = u + at, s = (u+v/2)t

Sub in for v

2ut + at^2/2 = s

s = ut + 0.5at^2

That wasn't so hard, was it?

 

[untouchablecuz]

note that acceleration is constant

edit: using calculus if anyones interested