Projectile Motion (1 Viewer)

[CM_Tutor]

I took g as g in all three

 

[nike33]

ok just asking in the 1st qn it canceled out, just making sure for the latter questions

 

[nike33]

id say im missing something but for 2ii...

V^2 = kh

maximum range: x = v^2 / g x = kh / g

ie max range is directly proportional to the height .:.

height = 4D ?/??

 

[CM_Tutor]

Re-read the question for the definition of h. At 4D above the ground, the range is 0.

 

[nike33]

oh hehe, i just took h as the height of the leak , ill try again tommorow maybe

 

[CM_Tutor]

Originally posted by nike33
oh hehe, i just took h as the height of the leak , ill try again tommorow maybe

That'd be physically unrealistic, and would make the question way too easy!

 

[nike33]

haha i was getting worried at your standards! something had to be wrong

 

[Xayma]

Is the dam wall straight or concave (looking straight across the water)? and why would the land be flat the natural sediments from falling in the dam would arrange itself to produce a banked style on the bottom.

Anyway... (I didnt prove all the s=ut+1/2at² etc since I havent done that in class yet)

Heres 2(a) (I bashed it, cause I couldnt think of an elegant way using Kinetic and potential energy) I will do 2(b) in a sec (although I have a feeling what the answer might be from logic)

And sorry about the formatting but I couldnt remember how to just do a html line so they are all < p > spaced.

And I had to put it in a .zip because they wont accept .html

 

[Xayma]

Ok 2(b) (it wouldnt let me put multiple attachements in the one post)

 

[Calculon]

I would try the cambridge questions but only got 4u cambridge

 

[CrashOveride]

(i) Prove that the jet can reach the wall above ground level if and only if V >(gc)1/2.

i))

i cant be bothered deriving formulas or watever...but maximum rangle v^2 = xg ..v = sqr(xg) taking it as positive

now let the base of the wall be the origon hence C(c,0)
now for the water to hit the wall at ground level exactly v = sqr(gc) as x = c

obviously now as v increases the distance increases. hence as
v = sqr(gc) hits the wall at ground level, if v > sqr(gc) it must hit the wall. and only if v > sqr(gc) as allready stated, this is the maximum range obtanible

Why do we say max range is needed here? Couldn't it just hit the bottom of the wall at a smaller range ? I had a similar question and assuming no max range thing, i got the condition: sin2@.U² > gd. So this would cater for all angles of projection? :S

Also the result U² = 4gd, i sub that in the c=Ucos@.t and we say 45 degrees because here max angle wud give us the highest point on wall? xcept i end up getting like 6d/8. If im missing something obviousy here, someone make me aware of my alarming decline of logic

 

[~*HSC 4 life*~]

I would try the cambridge questions but only got 4u cambridge

now i know what to get you for your birthday...


even though it's after the HSC lol

 

[idkkdi]

Nike33, it is good that you did prove that the shape on the wall is a parabola.

I took a different approach to solving this problem, which was:

Note that the distance travelled in the horizontal direction is kc, where 1 <= k <= 4, as 4c is the maximum range.
So, the time of flight is kc / [2 * sqrt(gc) * cos@] where @ is the angle of projection.
The height of impact, H, is: H = kctan@ - (k²c / 8) * sec²@.
dH/d@ = 4kcsec²@(4 - ktan@)
and it quickly follows that tan@ = 4 / k for a maximum height, and H_MAX = 2c - ck² / 8

When k = 1, H_MAX = 15c / 8 and when k = 4, H_MAX = 0, as expected.

If you take D as the distance along the base of the wall, with D = 0 when k = 1 - at the closest point to C - then Pythagoras' theorem shows that:
D² = k²c - c², and so k² = 1 + D² / c²
So, H_MAX = 2c - (c / 8) * (1 + D² / c²) = 15c / 8 - D² / 8c.

This is clearly a parabola, as required, and the area is
2 * int (from 0 to c * sqrt(15)) H_MAX dD = 5c² * sqrt(15) / 2, as required.

@CM_Tutor
"

If you take D as the distance along the base of the wall, with D = 0 when k = 1 - at the closest point to C - then Pythagoras' theorem shows that:
D² = k²c - c², and so k² = 1 + D² / c²
So, H_MAX = 2c - (c / 8) * (1 + D² / c²) = 15c / 8 - D² / 8c.

This is clearly a parabola, as required, and the area is
2 * int (from 0 to c * sqrt(15)) H_MAX dD = 5c² * sqrt(15) / 2, as required. "
Got confused this part.

My reasoning for the area involved, that the parabola is essentially a -x^2 scaled larger by 15/8C. And then i take an integral, but i seem to get a wrong answer.

1974 q10 btw.