Mathematics of Projectile Motion

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 
felt like I was going nowhere. Can you please help me?   

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
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