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### Mathematics of Projectile Motion

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Date: 7/7/96 at 20:56:27
From: Anonymous
Subject: Mathematics of Projectile Motion

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 10 m/s.  The stone reaches
double the height of the pole and, in its descent touches the bird.
Find the horizontal component of the velocity of the stone.

I tried to work out the velocity(v) of the stone (not vertical). I
found the maximum height equation in terms of v. I then said that the
pole is half the equation of the maximum height. Then I got stuck and
don't know what to do. It didn't matter what I did, I got it wrong and
```

```
Date: 7/8/96 at 19:44:56
From: Doctor Anthony
Subject: Re: Mathematics of Projectile Motion

These problems can be rather tricky with the number of equations that
can be written down. If you draw the parabolic arc representing the
path of the stone, this path must pass through the top of the pole
(let its coordinates there be (x1,h) with h = height of the pole) then
to the top of its flight with coordinates (x2,2h) and then to the
point where it brushes the bird at coordinates (x3,h).  Then we note
that if t1, t2, t3 are the times for the stone to reach x1, x2, x3
respectively, then t3 is also the time for the bird to travel a
distance (x3-x2) at 10 m/sec, that is 10*t3 = x3-x1  and this is the
equation I shall be using to find the horizontal velocity of the
stone.

For horizontal motion of the stone we have x = Vcos(E)*t  where E is
the angle of elevation and V is the velocity of projection. For
vertical motion  y = Vsin(E)*t - (1/2)gt^2

Time to highest point is given by the equation'v=u+at' which has v=0
at the highest point.  So 0 = Vsin(E) - gt  so t = Vsin(E)/g

y = 2h at the highest point, so we can write

2h = Vsin(E).Vsin(E)/g - (1/2)gV^2.sin^2(E)/g^2

2h = V^2.sin^2(E)/g - (1/2)V^2.sin^2(E)/g

= (1/2)V^2.sin^2(E)/g

h = (1/4)V^2.sin^2(E)/g

Now we can use this result to find the two times when the stone is at
height h, and hence find t1 and t3 and x1 and x3.

We have h = Vsin(E).t - (1/2)gt^2

(1/4)V^2.sin^2(E)/g = Vsin(E).t - (1/2)gt^2

(1/2)gt^2 - Vsin(E).t +(1/4)V^2sin^2(E)/g = 0

2g^2.t^2 - 4Vgsin(E).t + V^2sin^2(E) = 0

t = [4Vgsin(E) +or-sqrt{16V^2.g^2.sin^2(E) -
8V^2.g^2.sin^2(E)}]/4g^2

= [4Vgsin(E) +or-sqrt{8V^2.g^2.sin^2(E)}]/4g^2

= [2Vgsin(E) +or-Vgsin(E)sqrt(2)}]/2g^2

= Vsin(E)(2 +or-sqrt(2))/2g

So t1 = [Vsin(E)(2-sqrt(2)]2g    and t3 = [Vsin(E)(2+sqrt(2)]/2g

To get x3-x1 we multiply by Vcos(E)

So x3-x1 = [V^2.cos(E)sin(E){2+sqrt(2) - 2+sqrt(2)}]/2g

= [V^2.cos(E)sin(E){2sqrt(2)}]/2g

= [V^2.cos(E)sin(E).sqrt(2)]/g

Now equate this to 10*t3 = distance flown by bird

[V^2.cos(E)sin(E).sqrt(2)]/g = 10*[Vsin(E)(2+sqrt(2)]/2g

Vcos(E) = 5[2+sqrt(2)]/sqrt(2)

and so the horizontal velocity of the stone is 5[2+sqrt(2)]/sqrt(2)

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Physics/Chemistry

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