- #1

hwall95

- 16

- 0

## 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(∅) - gx

^{2}/2u

^{2}(tan

^{2}(∅)+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(∅) - gx

^{2}/2u

^{2}(tan

^{2}(∅)+1)

tan 48 = 1.11 u = 35 g = 9.8

y = 1.11x - 9.8x

_{2}/2450(2.2334)

= 1.11x -0.004x

^{2}*2.2334

= 1.11x - 0.00893x

^{2}

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 :)