Stones hitting Birds :D (1 Viewer)

[Smilebuffalo]

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.

 

[kwabon]

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]

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.

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

 

[gurmies]

http://img199.imageshack.us/img199/889/scan0002xo.jpg

 

[kwabon]

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

yes, i got that.

 

[Iruka]

My solution.

View attachment 19238

 

[gurmies]

My solution.

View attachment 19238

Nice

 

[azureus88]

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]

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)

 

[Aerath]

Sounds oddly familiar.....

 

[lolokay]

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
"

 

[David Obeid]

My solution.

View attachment 19238

What am I doing wrong? When I click on the link I just get a blank page?

 

[jskeza]

What am I doing wrong? When I click on the link I just get a blank page?

lol, maybe because the post is 10 years old and the attachment has been deleted or moved.