Projectile question - Fitzpatrick question!! (1 Viewer)

utanobeiiby · Jul 23, 2011

Q17 p124 : 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 10m/s. The stone reaches a max height that is twice the height of the pole and on its descent it touches the bird. Find horizontal component of velocity of the stone.

Ans from the book:12.1m/s

I've tried many ways but it seems there is something missing in the question and i haven't been able to come up with any numerical answer. So can someone please explain how to do this question??

hscishard · Jul 24, 2011

utanobeiiby said:
Q17 p124 : 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 10m/s. The stone reaches a max height that is twice the height of the pole and on its descent it touches the bird. Find horizontal component of velocity of the stone.

Ans from the book:12.1m/s

I've tried many ways but it seems there is something missing in the question and i haven't been able to come up with any numerical answer. So can someone please explain how to do this question??

I got the solution...I remember a thread made here aggeess ago about this question. I don't think anyone got the answer. At that time I didn't learn projectile motion but anyway I'll actually post the solution.

The birds path is the boldest line. Since the bird flys horizontally, the rock can only hit it again when it descends to a height of "h".

I have omitted some unnecessary working
$h=\frac{gt^2}{2}+Vtsin\alpha$
Solve it and you will get
$t=\frac{Vsin\alpha \pm \sqrt{(Vsin\alpha )^2 -2gh}}{g}$

The maximum height achieved is 2h.
$\frac{v^2sin^2\alpha}{2g} =2h$
$v^2sin^2\alpha =4gh$
$vsin\alpha =2\sqrt{gh}$

Therefore the whole thing would become
$t=\frac{2{\sqrt{gh}\pm \sqrt{2gh}}}{g}$
Therefore the x coodinates of the y=h are:
Let z=Vsin(@)
$x=\frac{2z\sqrt{gh}\pm z\sqrt{2gh}}{g}$
Then the distance between the two y=h is
$x=\frac{2z\sqrt{2gh}}{g}$
The bird travels this distance in
$time=\frac{z\sqrt{2gh}}{5g}$

The bird flew immediately when the stone was thrown and it was hit after flying for the time found. The horizontal distance of where it was hit is
$x=\frac{2z\sqrt{gh}+z\sqrt{2gh}}{g}$
(as this one is larger from the two)
horizontal speed is therefore
$speed=\frac{\frac{2z\sqrt{gh}+z\sqrt{2gh}}{g}}{\frac{v\sqrt{gh}}{5g}}$
Which simplfies to [10+5root2]/root2
=12.071m/s

hscishard · Jul 24, 2011

This took 35 minutes to type up, hence this is why I will never ever post a full solution again

utanobeiiby · Jul 26, 2011

wow thanks heaps! i was heading in that direction... but made some mistakes along the way so my g and h didnt cancels out >< anyways thanks for that! ur awesome !

hscishard · Jul 27, 2011

Mistakes in these questions are quite frustrating I know

Projectile question - Fitzpatrick question!! (1 Viewer)

utanobeiiby

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