3u Q. (1 Viewer)

[ND]

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 height double that of the pole and, in its descent, touches the bird. Find the horizontal component of the velocity at the stone.

This is from 3u Fitzpatrick. I couldn't do this at the start of the year (hopefully i'd be able to do it now), i think it's a nice question so i thought i'd post it.

 

[underthesun]

Do we take g as 9.8 or 10? Some textbooks tells you to take g as 10, I'm not sure about fitzpatrick..

edit : they give an answer in the back in exact numbers and not in terms of g right?

 

[stag_j]

we had a very similar question in our trial i think.
they put a trick in ours and said that the stone was launched at a height of 1m off the ground. and i think the pole was 6m high, so the trick was to see that there was only a 5m difference - you wouldnt believe how many ppl failed to see this.

in maths i think we just about always take g to be 10. it keeps the working much easier generally.

 

[Fosweb]

This is a HUGE question!!

Our teacher showed us this from another book also (maybe the 50 tips book, and it was rated 5 stars...)
We came to the conclusion, that in 3U (unless they wanted to be REALLY slack and make it an ultra hard qu 7) there would be some direction/show this is this etc.

Come on Mr Lee! Give us a solution (which i already have) without us having to call you!

 

[underthesun]

Actually, the solution seems quite short for my working, it makes me wonder whether I did it right or wrong..

by the way, is the answer 10 + 10(sqrt(2)) m.s^-1? The answer is independent of any values except for g..

edit : I misread the question, my answer is incorrect, so move on

I misread the question as if It reached it's maximum height on the same position as the pole, that is, it reached double the height of the pole AT the pole.. that makes the question easier i gues..

 

[ND]

Originally posted by stag_j
we had a very similar question in our trial i think.
they put a trick in ours and said that the stone was launched at a height of 1m off the ground. and i think the pole was 6m high, so the trick was to see that there was only a 5m difference - you wouldnt believe how many ppl failed to see this.

in maths i think we just about always take g to be 10. it keeps the working much easier generally.

I think it would be quite a bit easier if you know the height of the pole.

 

[ND]

Originally posted by underthesun
Actually, the solution seems quite short for my working, it makes me wonder whether I did it right or wrong..

by the way, is the answer 10 + 10(sqrt(2)) m.s^-1? The answer is independent of any values except for g..

The answer in the back is 12.1m/s. I'll have a go at it tonight.

 

[flyin']

Hey there. There is a really really elegant solution involvin' parabolas which one our Mathematics teachers showed us. (It's much shorter than the one you can find in the Ext1 50 Exam Tips which from memory has many lines of workin' and algebra.)

Essentially it sets out a generic parabola and works around it.

 

[Fosweb]

Can you post this solution please? Because looking back at the way our teacher showed us (from 50 tips) it really makes little sense now...

 

[ND]

Ok we have x=Vtcos@ ...[1] and y=-gt^2/2 + Vtsin@ ...[2].
Let pole have height h. Max height reached is 2h, and this happens when v=0:
dy/dt = -gt + Vsin@ = 0
t = Vsin@/g

sub in y=2h and t=Vsin@/g to [2]:
2h = -g(Vsin@/g)^2/2 + Vsin@(Vsin@/g)
h = (Vsin@)^2/4g

To find the times when the stone is at height h, sub y=h=(Vsin@)^2/4g into [2]:

(Vsin@)^2/4g = -gt^2/2 + Vtsin@
2g^2*t^2 - 4Vgsin@t + (Vsin@)^2 = 0
t = 4Vgsin@+-sqrt(16g^2(Vsin@)^2-8g^2(Vsin@)^2)/4g^2
= 2Vsin@ +- sqrt2*Vsin@/2g
= Vsin@(2+-sqrt2)/2g

So the first time the stone is at h, is t_1=Vsin@(2-sqrt2)/2g, and the second time is t_2=Vsin@(2+sqrt(2))/2g. Now to get the distance between the pole and the place where the bird was hit, we sub into [1]. Let x_1 and x_2 be the positions the stone is at after t_1 and t_2 resp.
t_2-t_1 = Vsin@(2+sqrt(2))/2g - Vsin@(2-sqrt2)/2g
x_2-x_1 = Vcos@(Vsin@(2+sqrt(2))/2g - Vsin@(2-sqrt2)/2g) from [1]
= sqrt2*V^2*cos@sin@ ...[3]

Now, the time t_2 for the stone to reach x_2, is also the time for the bird to reach x_2 travelling at 10m/s from x_1:
x_2-x_1 = 10*t_2
subbing into [3] gives:
sqrt2*V^2*cos@sin@ = 10*t_2
now from before, t_2=Vsin@(2+sqrt(2))/2g, so:
sqrt2*V^2*cos@sin@ = 10Vsin@(2+sqrt(2))/2g
Vcos@ = (10+5sqrt2)/sqrt2

 

[ND]

I've got another solution (maybe the one flyin' was talking about?):

Ok consider a parabola with height 2h and let the roots be where the stone leaves from and touch (or would touch) the ground. Now solving this simultaneuously with line y=h, to make the roots the points where the stone touches the pole and then the bird. Now remember time t_1 for the stone to hit the bird is also the time it takes the bird to travel between these roots at 10m/s. So let the roots be 5t_1 and -5t_1. This parabola would have eqn:

y = -x^2 + 25t_1^2
when x = 0, y=h:
h = 25t_1^2

Now when y=-h=-25t_1^2:
x^2=50t_1^2
x = +-5sqrt2*t_1
Now x=-5sqrt2*t_1 will be where the stone was launched from (assume it was launched left to right), so the total distance travelled by the stone before it hits the bird is:
5*t_1*(1+sqrt2)

Now displacement = velocity x time
so 5*t_1*(1+sqrt2) = t_1*v
v = 5*(1+sqrt2)