Mathematics of Projectile MotionDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
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