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?"
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,