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Find where and when the trajectory angle is 10 degrees?

  • Thread starter hwall95
  • Start date
  • #1
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Homework Statement


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?"


Homework 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?

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 :)
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
7
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.
 
  • #3
16
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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.
Okay thanks, that makes a lot of sense actually. Haha thanks for your help :)
 
  • #4
1,065
10
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
 
Last edited:
  • #5
16
0
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
Hahaha wow thats even simpler, haha i guess it saves messing around with the path equation then, thanks heaps :)
 

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