Stones hitting Birds :D (1 Viewer)

Smilebuffalo

Member
Joined
Jun 9, 2008
Messages
89
Location
Fairfield West
Gender
Male
HSC
2010
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 descent, touches the bird. Find the horizontal component of the velocity of the stone.



:bomb:
 

kwabon

Banned
Joined
May 26, 2008
Messages
646
Location
right behind you, mate
Gender
Male
HSC
2009
i have a way to do that question, but it has a whole heap of algebra in it.
sooo unfortunately i cbf typing it up,

so lets just wait for an easier way :)

i will type it up, if no one answers this question by the end of tomorrow, which is most unlikely.
 

Drongoski

Well-Known Member
Joined
Feb 22, 2009
Messages
4,238
Gender
Male
HSC
N/A
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 descent, touches the bird. Find the horizontal component of the velocity of the stone.



:bomb:
is answer: 5(1 + sqrt(2) ) m/s ??
 

azureus88

Member
Joined
Jul 9, 2007
Messages
278
Gender
Male
HSC
2009
Alternatively,

Consider [maths]y=-kx^2[/maths] for [maths]-2h\leq y\leq 0[/maths]

when [maths]y=-h,x=\pm \sqrt{\frac{h}{k}}[/maths]

when [maths]y=-2h,x=\pm \sqrt{\frac{2h}{k}}[/maths]

D1 (distance travelled by stone) [maths]=\sqrt{\frac{2h}{k}}+\sqrt{\frac{h}{k}}[/maths]

D2 (distance travelled by bird) [maths]=2\sqrt{\frac{h}{k}}[/maths]

[maths]\frac{D_1}{V}=\frac{D_2}{10}\\V=10\frac{D_1}{D_2}\\=10\left [ \frac{\sqrt{\frac{h}{k}}(\sqrt{2}+1)}{2\sqrt{\frac{h}{k}}} \right ]\\=5(\sqrt{2}+1)\\=12.1[/maths]
 

untouchablecuz

Active Member
Joined
Mar 25, 2008
Messages
1,693
Gender
Male
HSC
2009
Alternatively,

Consider [maths]y=-kx^2[/maths] for [maths]-2h\leq y\leq 0[/maths]

when [maths]y=-h,x=\pm \sqrt{\frac{h}{k}}[/maths]

when [maths]y=-2h,x=\pm \sqrt{\frac{2h}{k}}[/maths]

D1 (distance travelled by stone) [maths]=\sqrt{\frac{2h}{k}}+\sqrt{\frac{h}{k}}[/maths]

D2 (distance travelled by bird) [maths]=2\sqrt{\frac{h}{k}}[/maths]

[maths]\frac{D_1}{V}=\frac{D_2}{10}\\V=10\frac{D_1}{D_2}\\=10\left [ \frac{\sqrt{\frac{h}{k}}(\sqrt{2}+1)}{2\sqrt{\frac{h}{k}}} \right ]\\=5(\sqrt{2}+1)\\=12.1[/maths]
sexy (y)
 

lolokay

Active Member
Joined
Mar 21, 2008
Messages
1,015
Gender
Undisclosed
HSC
2009
Sounds oddly familiar.....
indeed it does..

my solution was this (however it's not really the projectile motion approach)
"
say the projectory of the stone is represented by y=-Cx2, so that the origin is the maximum point/height
if d is the distance the stone travels to get from a height of 2p (max) to p, then p = Cd2.
The distance, D, that the stone travels in going from 2p to the ground would then be rt2 d [2p = CD2 = 2Cd2 -> D = rt2 D] (same distance to get from ground to 2p)
So the total horizontal distance travelled by the stone is d + rt2 d, whereas the total distance travelled by the bird is 2d

They cover each of these distance in the same time, so you can then show that vstone/dstone = vbird/dbird

vstone = 10(rt2 + 1)/2 m/s
= 12.07m/s
"
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top