hard question in exercises

[cotfw]

I was working through my exercises and they included this one that is very unlike the others.

"A particle is projected at an angle θ with initial velocity 'u'. Its horizontal displacement (x) and height (y) are connected by the equation:

y = x tanθ - gx²/2u²cos²θ

What is the maximum height to which the particle rises, and how far has it then travelled horizontally? (g is the acceleration due to gravity, a constant.)"

I thought I'd find f'(x) first, and I ended up with tanθ - 2g2u^-2cos^-2θx.

Then I thought I should find which value of x makes the above equal zero:

tanθ/2g2u^-2cos^-2θ = x

At this point, I lose faith that I'm on the right path. I feel confused about what i should do.

 

[Ishuda]

Yes, you are on the right path.

To make the equation easier to handle (at least for me), let
a = tan(\(\displaystyle \theta\))
b = \(\displaystyle \frac{g}{2\space u\space cos^2(\theta)}\)
Then
y = a x - b x²
and
y' = a - 2 b x
So, as you pointed out, when
x = x₁ = \(\displaystyle \frac{a}{2 b}\)
y' is zero. Now you need to check y'' to see if y is a max or min or possibly neither at x=x₁.

Hint: You will also have to evaluate y at x = x₁