hrm, projectile motion (1 Viewer)

[Grey Council]

hrm, teaching myself projectile motion today.

anyway, here is a question i can't figure out. Actually, I CAN, but I'm puzzled about one particular part of the question:
A stone is projected from the top of a cliff 5g(sqrt 2) m high. The horizontal and vertical components of the velocity are initially
2g(sqrt 3) m/s and 2g m/s respectively.

At what angle does the stone hit the sea?

i've done a fair bit of stuff, and this is what I have:
at the instant of hitting the sea, the stone's velocity is (sqrt 21)g. Its horizontal velocity is 2(sqrt 3)g and its vertical velocity is -3g.

the angle of hitting the sea is therefore (by arctan i mean inverse tan)
arctan -3/(2 sqrt 3) = arctan (-sqrt3) / 2
which is more or less 40 degrees. The answer is 180-40. why? why do we take it away from 180? which angle do we want if the question asks that?

 

[Xayma]

I think it wants the angle in relation to the positive x-axis. As that is how angles between the x-axis and lines are generally measured.

 

[Grey Council]

but that gives me 40, not 139. huh, wtf, i think i'm drawing the diagram wrong. hmmm..

 

[KeypadSDM]

Dude, the answer is more or less -40 degrees

Thus the answer, since angles have to be positive, is 180 more than that, 139.106605.... degrees
or 139`6'24"

 

[xiao1985]

Originally posted by KeypadSDM
Dude, the answer is more or less -40 degrees

Thus the answer, since angles have to be positive, is 180 more than that, 139.106605.... degrees
or 139`6'24"

lol,z or more or less 40 degrees below the horizon...

 

[KeypadSDM]

Originally posted by xiao1985
lol,z or more or less 40 degrees below the horizon...

Which is a legit answer, as it is technically the angle it makes with the 'positive x axis', but the positive angles are more generally accepted. I hope I'm not ranting.

 

[Grey Council]

as i said, more or less. take or give a negative sign here or there etc.

hehe, but you see what I meant.

thanks for the reply. I think I understand now. Add 180, and since its negative it becomes 180-40 (or whatever the heck it is 39.takestoolongtotypeup).

I see

 

[Grey Council]

WTF< i am very confused now.

A ball is thrown at a velocity of 14.7 m/sec. Find the greatest distance to which it could be projected and the height to which it would rise.
Yes, I know, this is easy. Then why the hell do i keep getting it wrong? I've tried to derive the answer, tried using a formula, still can't do. humph. This is what i've got so far (skipping most of the working)

x' = 14.7/sqrt2
y' = 14.7/sqrt2 - gt

.'. x = 14.7t/sqrt2 and
y = 14.7t / sqrt2 - 1/2.g.t^2

for max range y=0 with theta 45 degrees etc. sub in t value into x:
t=29.4/gsqrt2

value of max range =
14.7/sqrt2 x 29.4/gsqrt2
= 432.18/2g

whatd i do wrong? :S maybe the answers are wrong? somehow that doesn't sound convincing. humpho

 

[Xayma]

Can you just assume that the best angle is 45 degrees or do you have to prove it.

If you can just assume:

Ok max height:

KE₁+PE₁=KE₂+PE₂
m(.5v²₁1)+0=m(.5v²₂+gh)
.5x14.7²=(.5x[14.7/sqrt2]²)+gh
h=54.0225/g metres

Is that the answer or not...

*Xayma accepts no responsibility for using the above formulas in a test, they are outside the scope of the syllabus and you may need to prove them to use them.

 

[Grey Council]

nope, not according to the answers. bl, try again. and where are the formulas from?

 

[Xayma]

That should be the height when thrown at a 45 degree angle. Those are energy formulas from Engineering Studies, due to the conservation of energy the Kinetic Energy+Potential Energy at the start=Kinetic Energy+Potential Energy at the end

KE=1/2mv²
PE=mgh

You can take out the factor of m at the start.
At Max height the only velocity is in the x-direction ie 14.7sin45 degrees=14.7/sqrt2.

Following the formulas gives you my answer.

If you can get the max height it is easy to get the time taken as you can use r=ut+1/2at² then twice it to get the total time then times it by x (yeah it can be easier to find that, but just for height you dont need the time).

Btw what is the answer?

 

[KeypadSDM]

Originally posted by Xayma
Can you just assume that the best angle is 45 degrees or do you have to prove it.

That cost me 1st in the HSC. Don't EVER EVER do that.

 

[Xayma]

I guessed that but I dont know how to prove it at the moment.

 

[Grey Council]

lol. I knew keypad would say that.

anyway, can you help? blah to you Keypad. You could have typed up the answer in the time it took you to write that advice. humph

the answer is supposedly (from what I can read of it anyway, this is word for word):

4.41cm, 4.41cm (note theta = 45 degrees)

WTF, i don't understand how that is possible, the velocity is in metres per second, i don't get it.

and the proving of 45 degrees is quite easy. get max range, you'll have a sin2theta in there somewhere. Sin theta's maximum value is 1, therefore theta is 45. or something like that. Its not hard, just do it the normal way. (lol at keypad. ahaha, IGGIES DOWN THE DRAIN )

EDIT:
I've just had a revelation.
maybe the answer at the back of the book IS wrong.

i was like, stunned. maybe the answer is wrong. In fact, i'm sure it is wrong. The initial velocity is in metres per second. hrmm, stoopid coroneos book. Thats two wrong answers in two questions. that is dodgy.
*hits himself for not even thinking about the answer*
oh god.

 

[KeypadSDM]

Originally posted by Grey Council
(lol at keypad. ahaha, IGGIES DOWN THE DRAIN

Yeah well at least JR hasn't got anything on us.

 

[Xayma]

Originally posted by Grey Council
maybe the answer at the back of the book IS wrong.

Thats alot of frictional resistance in the air if only goes that far.

 

[Grey Council]

lol, so now what? I skip this question? I think i'll do that.

aren't I smart?

Originally posted by KeypadSDM
Yeah well at least JR hasn't got anything on us.

hah! obviously not. James Ruse sucks. ^_^ its only good cause all the nerds go there, and it has a reputation, which leads even more nerds there, which means it has an even better reputation ad infinitum.

Iggies have (or should logically have) better teachers and better facilities. In fact, i've been to iggies to play sport, and I KNOW the facilities there are better. heheh, the students DO pay like $10000 a year. hrm, correct me on this, i've always been curious as to the exact figure. ah well, public schools rock. (in a way and to a point)

 

[freaking_out]

Originally posted by Grey Council
.... ah well, public schools rock. (in a way and to a point)

yeah well- if only public schools can have the same callibre of teachers as there is in private skools!

 

[KeypadSDM]

Originally posted by freaking_out
yeah well- if only public schools can have the same callibre of teachers as there is in private skools!

Yeah, well Private Schools can also be sucky.

I was lucky my maths teacher was awesome.

 

[freaking_out]

Originally posted by KeypadSDM
Yeah, well Private Schools can also be sucky.

I was lucky my maths teacher was awesome.

no- they'll always have the top quality teachers! but they can b sucky in terms of cost though.