Question

A projectile thrown from a point P moves in such a way that its distance from P is always increasing. Find the maximum angle above the horizontal with which the projectile could have been thrown. Ignore air resistance.

Answer 1

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!!!