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Bird and stone = hard? (1 Viewer)

xionethic

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I need help with this Projectile Motion question:

A stone is thrown so that it will hit a bird at the top of a pole. However, at the instant the stone is thrown, the bird flies away in a horizontal straight line at a speed of 10ms^-1. The stone reaches a height double that of the pole and, in its decent, touches the bird.
Find the horizontal component of the velocityu of the stone.
 

YannY

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xionethic said:
I need help with this Projectile Motion question:

A stone is thrown so that it will hit a bird at the top of a pole. However, at the instant the stone is thrown, the bird flies away in a horizontal straight line at a speed of 10ms^-1. The stone reaches a height double that of the pole and, in its decent, touches the bird.
Find the horizontal component of the velocityu of the stone.
Considering only the horizontal components
let the distance from the pole and the thrower be a. let the time it takes to hit the bird be t.

therefore in t seconds the bird travels10t m and thus the stone travelled a+10tm

v=d/t
therefore v=a/t+10ms
 

YannY

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xionethic said:
The answer is 12.1m/s oO, err... what did you get? lol
you cant work out that because you didnt give me the whole question.... how would you get 12.1m/s when you were only given one value? so what miraculously 10/ms just turned into 12.1m/s?
 

jcurry

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how far way from the pole was he to start off with, did the bird fly towards him or away???
not enough information
 

xionethic

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YannY said:
you cant work out that because you didnt give me the whole question.... how would you get 12.1m/s when you were only given one value? so what miraculously 10/ms just turned into 12.1m/s?
That is the whole question. Honest to god i can photocopy u the page and link you it. My maths teacher said you can get it from first principle, working out is about 2-3 pages, his a capable maths teacher, teaching me in 4unit atm. He hasn't told me his progress due to the fact that he hasnt had time to sit with the question

10 didn't just 'turn' into 12.1, it's a constant. The horizontal velocity of the rock is 12.1 m/s. This can be taken by the relative movement of both the rock and the bird because they both collide at a particular point. If the time is the same, the horizontal displacement of the collision is the further for the rock than the bird, from their respective origins, we know that logically, the horizontal speed of the rock is faster. But the question is what variables and formulas do i use to derive the answer?
 
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xionethic

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And, the question is possible to do, a friend of mine got a value out althought it was incorrect. The answer derives from using a few variables.
 

YannY

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okay sorry im not really that great at maths. Can you get the solution from your teacher and show it to me? this question is quite interesting.
 

SpinCobra

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I agree. Post it up.

One day this question is gonna fuck us over in the HSC. Haha.
 

SpinCobra

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I still dont see it. Or maybe I've got something on my mind and Its wrong.

Height of the pole is variable, right?

Wouldnt the horizontal component of the rock vary as it was lower or higher?

Maybe its cause I'm sleepy. I dunno. But it just doesnt make sense at the moment.
 

xionethic

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SpinCobra said:
I still dont see it. Or maybe I've got something on my mind and Its wrong.

Height of the pole is variable, right?

Wouldnt the horizontal component of the rock vary as it was lower or higher?

Maybe its cause I'm sleepy. I dunno. But it just doesnt make sense at the moment.
Horizontal components never vary? they are constant, same with downwards component which is gravity. Horizontal and vertical component do not effect each other and can be treated as two seperate terms.
 

xionethic

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the rock does not reach double the height exactly on top of the pole, because it is aimed at the pole, it will draw a parabolic motion and the maximum height is after the pole, and hitting the bird at the same height as the pole. sorry i mean touching, test writters don't like to be cruel :D
 

SpinCobra

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Yes but. Say h was lower. Therefore max height is lowered making the parabola flatter. Which means that the horizontal component will have to increase to reach the bird?

Okay I'mma head to sleep. This makes no sense at all right now.
 

xionethic

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SpinCobra said:
Yes but. Say h was lower. Therefore max height is lowered making the parabola flatter. Which means that the horizontal component will have to increase to reach the bird?

Okay I'mma head to sleep. This makes no sense at all right now.
oh you're talking about that, yea the birds horizontal component is different to the rocks. They are two different events. Is that what you mean?
 

xionethic

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lol im studying english and doing maths i cant do this its too conflicting xD ill post it up if i ever get the solution.
 

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Has anybody solved this freaking question???

I've just tried to do it and i can't get it out...i'd be interested in seeing the solution if anyones been able to do it.
 

lolokay

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t1 = time for rock to reach pole
t2 = time from pole to max height
t3 = time from max height to pole height, which is time it hits bird, = t2
t.t = total time = t(1+2+3)
p = pole height
r = square root pole height
d = distance from initial point of rock to pole (horizontal distance)

Using equations v^2 = u^ + 2as and v = u + at you get approximate values of t(1+2) = 6.26r/9.8 t3 = 4.43r/9.8, so t1 = 1.83r/9.8, t2 = 4.43r/9.8, t.t = 10.69r/9.8.

0 = u^2 + 2*-9.8*2p
u ^2 = 39.2p
u = 6.26r
0 = 6.26r - 9.8t
t = 6.26r/9.8
this is section t(1+2)

v^2 = 0 + 2*-9.8*-p
v^2 = 19.6p
v = 4.43r
4.43r = 0 + 9.8t
t = 4.43r/9.8
this is section t3


The distance the bird has travelled is 10t.t = (106.9r)/9.8.
The horizontal distance the stone has to travel is (106.9r)/9.8 + d.
The horizontal velocity is constant, so v = d/t1: d = (1.83r*v)/9.8

horizontal velocity = horizontal distance / time
v = (106.9r/9.8 + 1.83rv/9.8)/(10.69r/9.8)
10.69rv = 106.9r + 1.83rv
8.86rv = 106.9r
v = 12.065 m/s

since the values were approximations it probably is closer to 12.1 (the given answer)

PS diagrams really help :p
 
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paddy@2011

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xionethic said:
I need help with this Projectile Motion question:

A stone is thrown so that it will hit a bird at the top of a pole. However, at the instant the stone is thrown, the bird flies away in a horizontal straight line at a speed of 10ms^-1. The stone reaches a height double that of the pole and, in its decent, touches the bird.
Find the horizontal component of the velocityu of the stone.

here is a worked solution in the attachment.
 

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