Answer
70.52°
Explanation
The distance between projectile's position and it's starting point at any time is given by the relation
r² = x² + y²
where x = horizontal distance covered and y = vertical distance covered
According to projectile motion the horizontal displacement is given by
x = v(x)t = v cos(θ) t
Also the vertical component is given by
y = v(y) t - 0.5gt² = v sin(θ) t - 0.5gt²
Substituting the x and y values into the r-equation yields,
r² = (v cos(θ) t)² + (v sin(θ) t - 0.5gt²)²
r² = v²(cos²(θ))t² + v²(sin²(θ))t² – (vg sin(θ))t³+ 0.25 g²(t^4)
r² = v²t² (cos²(θ)+ sin²(θ)) – (vg sin(θ))t³ + 0.25 g²(t^4)
r² = v²t² – (vg sin(θ))t³ + 0.25 g²(t^4)
Differentiate r with respect to t
r(dr/dt) = 2v²t - 3vg sin(θ)t² + g²t³
At maximum angle the projectile could have been thrown above the horizontal, dr/dt = 0
2v²t - 3vg sin(θ)t² + g²t³ = 0
Divide through by t
2v² - 3vg sin(θ)t + g²t² = 0
g²t² - 3vg sin(θ)t + 2v² = 0
This can be solved using the general law for quadratic equations
(-b ± √(b² - 4ac))2a
a = g², b = -3vg sin(θ) c = 2v²
t = ((3vg sin(θ)) ± √(9v²g²sin²(θ) - 8g²v²))/2g²
This equation makes sense when the value under the square root is positive, that is, the square root exists.
9v²g²sin²(θ) - 8g²v² > 0
9sin²(θ) - 8 > 0
Meaning sin²(θ) = 8/9
Sin θ = (2√2)/3
θ = 70.52°
QED!!!
