Projectile Motion (1 Viewer)

[CM_Tutor]

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.

 

[~*HSC 4 life*~]

Originally posted by untamedanimal
You didnt chose physics solely to avoid projectile motion?

nah i didn't choose physics because i hated it all (well at the time) i think i might've liked it now

 

[nike33]

ahh yes nice CM thats alot better than my unelegant 1 1/2 page algebra bash

 

[Grey Council]

Another question for your doing pleasure.

A ball is dropped from a height of 49 meters above the ground. The height of the ball at time t is h(t)=49-4.9t2 meters. A light which is also 49 meters above the ground is 10 meters to the left of the ball's original position. As the ball drops, the shadow of the ball caused by the light moves across the ground. How fast is the shadow moving one second after the ball is dropped?

lol, I shouldn't be helping you guys, i'm competing with you. ^_^

 

[nike33]

can you clarify this ? 49-4.9t2 t^2 ? and im assuming 4.9 not 4x9

 

[Grey Council]

my apologies.

A ball is dropped from a height of 49 meters above the ground. The height of the ball at time t is h(t) = 49 - 4.9t² meters. A light which is also 49 meters above the ground is 10 meters to the left of the ball's original position. As the ball drops, the shadow of the ball caused by the light moves across the ground. How fast is the shadow moving one second after the ball is dropped?

yes, 4.9, not 4 x 9.

which school do you go to?

EDIT:
lol, these are 4u projectile motion questions, we should not be posting it here. Some people might be a tad bit discouraged. ^_^

 

[nike33]

haha i think i got it out, can u pm or post an answer (cbf typing 2pages out for nothing) would like to see CMs answer for this, like the other qn tho

 

[Grey Council]

lol, i'm afraid i have no solutions to these questions I post up. ^_^ no answers or solutions, I just collect any good/interesting questions I see. I have quite a compilation now, for heaps of topics.

And you haven't answered my question. Which school you attend? You can PM me, if you want. ^_^

 

[Xayma]

Hmm I havent done trigometric differentiation etc, but if you can find the rate of change of the angle between a beam of light and the ball (which Im trying to figure out a way to do it the moment only using windows calc cause mine is at my mums).

Assuming that the distance from the distance from the final resting place is d then d=49/tan@.

Now if you can find the rate of change of the angle I should be able to help you some more.

 

[CM_Tutor]

Nike33, I get that the shadow is moving to the left at 200 ms^-1 one second after the ball is dropped. Can you confirm this answer, and if you can I'll post up a really nice solution.

 

[nike33]

oik here goes...

notice similar triangles...

x/10 = (49 - 4.9t^2) / 4.9t^2

find dx / dt = [(4.9t^2)(-98t) - (490 - 49t^2)(9.8t)] / (4.9t^2)
sub in t = 1 and dx/dt = -200 ie v = 200m/s to the left

 

[nike33]

my other way i did it was an algebra bash, ended up with 200 but took 3-4 pages

 

[nike33]

alternativly (a 3rd way haha!)

dh/dt = -9.8t

then you get dx/dh = 20 20/49 (20.408...) at t = 1 and dh/dt = -9.8

now dx / dt = dh/dt * dx / dh
= (-9.8) * (20 20/49)
= -200

!!

 

[Calculon]

Originally posted by Grey Council
mate, i dunno about "i can't wait till 4u mechanics". Are you insane? wtf, i dunno how smart you are, but its pretty difficult to do the questions in the 3u cambridge.

And if you do the extension question in the cambridge book in the projectile motions chapter, i'll give you $10.

Its called sarcasm

 

[nike33]

grey u said that other projectile motion qn was in cambridge 3u....$10

 

[CM_Tutor]

Yeah, I did a similar triangles way too, but I had H / (D - 10) = 49 / D, where H = 49 - 4.9t², and D is measured along the ground from directly below the light.
So, DH = 49D - 490
Differentiating implicitly with respect to t, I got D * dH/dt + H * dD/dt = 49 dD/dt
which simplifies to dD/dt = -2D / t

Then, at t = 1 s, H = 44.1 m and D = 100 m, and so shadow is moving left at 200 ms^-1

 

[CM_Tutor]

Some more good projectile questions - note that these are pretty hard for Extn 1 students, and I can break them into more HSC-style structured versions if people would like. The would only be asked as they stand by an Extn 2 examiner in a bad mood, IMO:

1. A projectile is fired at a bird sitting on top of a pole. At the instant the projectile is fired, the bird flies directly away in a horizontal path at a constant speed. The projectile reaches a maximum height which is twice that of the pole, and hits the bird on the way down. Show that the ratio of the horizontal component of the velocity of projection of the projectile to the speed of the bird is (sqrt(2) + 1) : 2.

2. A dam wall is holding back a body of water that is 4D units deep. If a leak occurs in the wall, water escapes horizontally from the leak at a velocity V determined by V² = kh, where k is a positive constant and h is the height of water above the leak. The ground is level on either side of the dam wall.

(a) Show that leaks occuring D units from the ground and D units from the water surface will land in the same place.

(b) Find the height above the ground for a leak to land as far as possible from the base of the dam.

3. A man is standing on a hill which slopes at an angle @ to the horizobtal. He throws two balls, both at the same speed and both at 45 to the horizontal, but one up the hill and the other down the hill. If the range along the ground of the ball thrown down the hill is twice that of the ball thrown up the hill, show that @ = tan^-1(1 / 3).

 

[Grey Council]

Originally posted by nike33
grey u said that other projectile motion qn was in cambridge 3u....$10

hah! wasn't meant for you. Was meant for calculon, who I know is a dumbarse.

lmao at the bird question.
I got turtle to explain that to me. ^__^ i still have the proof lying carefully stored away.

 

[nike33]

hehe 1) works out nicely (damn projectile motion will be boring when we do it at school after these questions )

 

[nike33]

also are we taking g as 9.8 or 10 in these questions