Projectile Motion

Let $v$ be the velocity with which the particle had been projected and $\theta$ is the angle of projection. At time $t$ , image

Vertical component of velocity = $vsin \theta$

Horizontal component of velocity = $vcos \theta$

Vertical displacement of the projectile= $vsin \theta t - \frac{1}{2}gt^2$ ( $g$ only acts along vertical direction)

Horizontal displacement of the projectile = $vcos \theta t$

So, net displacement = $\sqrt{(vsin \theta t - \frac{1}{2}gt^2)^2 + (vcos \theta t)^2}$

Now, $f(t) = (vsin \theta t - \frac{1}{2}gt^2)^2 + (vcos \theta t)^2$ should be an increasing function.

So, $f'(t) \geq 0$

$f(t) = (v^2 sin^2 \theta t^2 +\frac{1}{4}g^2 t^4 -vsin \theta g t^3) + (v^2 cos^2 \theta t^2)$

$\implies f(t) = (v^2 t^2 +\frac{1}{4}g^2 t^4 -vsin \theta g t^3)$

$\implies f'(t)= 2v^2 t + g^2 t^3 -3vsin \theta gt^2$ ( since $\theta$ is constant)

Now, $f'(t) \geq 0$

$\implies g^2 t^3 -3vsin \theta gt^2 + 2v^2 t \geq 0$

$\implies g^2 t^2 -3vsin \theta gt + 2v^2 \geq 0$ (as $t \geq 0$ always)

It is a quadratic equation which has imaginary roots or real and equal roots.

Therefore, the $Discriminant \leq 0$

$\implies (3vsin \theta g)^2 - 4(g^2)(2v^2) \leq 0 \implies sin^2 \theta \leq \frac{8}{9}$

$\implies sin \theta \leq \frac{2 \sqrt{2}}{3}$ (considering only positive value)

$\implies \theta \leq 70.52$

Another solution can be, which does not use calculus is: The distance will stop increasing when the angle between the position vector of projectile at any point during flight with its velocity vector is 90 or less, as long as this angle is greater than 90 it will always move away from the point of projection so all we need to do is to make dot product of those two vectors and the condition at which angle is 90 degree gives us the required angle, Position vector = (u(\cos \theta) t)x + (u(\sin \theta) t-0.5 g t^2)y And velocity vector is = (u\cos \theta)x + (u(\sin \theta-g*t)y When you take dot product of these two vectors you gete same condition as you would have got by using calculus which is-

for angle to be 90 degree 2 u^2-3 (u(\sin \theta) g t+g^2*t^2 is greater than or equal to zero you get same answer sorry for my presentation above, but the method is finally what matters

A similar problem has already been posed earlier Always moving away I found it later in the part "More problems to try" and it was posed for Level 5.Check out Jatin Yadav's solution.

Arcsin(sqrt(8/9))