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Projectile Motion Problem |
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| Sep15-12, 08:05 PM | #1 |
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Projectile Motion Problem
1. The problem statement, all variables and given/known data
Okay, this is a projectile motion problem(No Drag) where a rock is thrown as 35m/s at an angle of 48 degrees. The question that im struggling with is: "Find where and when the trajectory angle is 10 degrees?" 2. Relevant equations The Equation that my teacher was used was that he used the path equation: y = xtan(∅) - gx2/2u2 (tan2(∅)+1) He then derived it and let it equal tan 10. However my main problem is that i dont understand why he derived it and then let it equal tan 10. Because the path equation depicts motion of projectile in relation to it y and x motion, thus if you derive it you would get the velocity motion which would not be useful for this application right? 3. The attempt at a solution This is my teacher's solution: y = xtan(∅) - gx2/2u2 (tan2(∅)+1) tan 48 = 1.11 u = 35 g = 9.8 y = 1.11x - 9.8x2/2450(2.2334) = 1.11x -0.004x2*2.2334 = 1.11x - 0.00893x2 dy/dx = 1.11 - 0.01763x tan 10 = 0.1763 = 1.11 - 0.01787x 0.01787x = 1.11-0.1763 x = 52.248m Ive tried using a range of equation, but they all rely on the initial angle, thus any attempt to sub 11° kinda wrecked the whole equation. Anyways if anyone could help me understand why the derivative of the path equation can work out when and where the trajectory angle is 11° it would be much appreciated, Cheers :) |
| Sep15-12, 08:56 PM | #2 |
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Recognitions:
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In the projectile motion, at any instant, dx/dt is the velocity in the horizontal direction and dy/dt is the velocity in the vertical direction. The direction of the velocity at that instant is given by tanθ = vy/ vx = dy/dx. That is what your teacher did.
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| Sep15-12, 09:05 PM | #3 |
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| Sep15-12, 09:52 PM | #4 |
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Projectile Motion Problem
I do it this way, with simple equation applied.
At 10° trajectory angle, Tan(10°)=vy/vx vy=35.Cos48°.Tan10°=4.13m/s t=(35.Sin48°-4.13)/9.8=2.23s x=2.23 x 35.Cos48°=52.22m |
| Sep16-12, 02:18 AM | #5 |
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