What is the physical explanation of why a $45°$ projectile angle gives the absolute maximum range? [closed]

Question

Assuming, start and end positions are at the same height and there is no air drag.

How gravity's role plays out during flight of projectile?

Answer 1

We need no calculus to prove this. Only conservation of energy and a bit of trigonometry.

A projectile motion without friction consists of two components: a uniform motion in the horizontal direction and a uniformly accelerated motion in the vertical direction.

The motion goes on until the projectile falls to the ground. If the vertical velocity is $v_y$ at the start, then when the ball returns to the ground it must have vertical velocity $-v_y$ due to conservation of energy (the total velocity must have the same absolute value, which is only possible if the vertical velocity at the end is $\pm$ the velocity at the beginning, given that the horizontal velocity doesn't change).

Since acceleration is $a=g$ (throughout I take $g$ to be negative), it takes a time of $\Delta t=\frac{\Delta v}{a}=\frac{-2v_y}{g}$ for the projectile to fall to the ground. In this time it goes $\Delta s_x=v_x\Delta t=\frac{-2v_xv_y}{g}$ to the right.

Now we also have that if total velocity has a modulus of $v$, then the components in the horizontal and vertical directions are $v_x=v\cos\alpha,v_y=\sin\alpha$, where $\alpha$ is the angle at which the projectile is being shot. We get

$$\Delta s_x=-2v^2\frac{\sin(\alpha)\cos(\alpha)}{g}.$$

Now here comes a bit of math slightly removed from the physical side of things: using the double angle formula, $\sin(\alpha)\cos(\alpha)=\frac{\sin(2\alpha)}{2}$. So we get

$$\Delta s_x=-v^2\frac{\sin(2\alpha)}{g}.$$

The sine is known to be maximal if its argument is $90^\circ$ (immediately obvious from the unit circle definition) so $2\alpha=90^\circ$ and thus $\alpha=45^\circ$.

Answer 2

The acceleration is the second derivative of the displacement:

$$a(t) = \frac{d^2}{dt^2} s(t)$$

If the acceleration is constant $a(t) = A$ then velocity is

$$\frac{d}{dt} s(t) = v(t) = \int_{0}^{t} A d\tau = A \cdot t + v_0$$

where $v_0$ is the initial velocity. The displacement is now defined as:

$$s(t) = \int_{0}^{t} (A \cdot \tau + v_0) d\tau = \frac{1}{2} A \cdot t^2 + v_0 \cdot t + s_0$$

where $s_0$ is the initial displacement. These are well-known equations for displacement and velocity with constant acceleration, you can find them in any basic physics textbook.

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In your example you have two axes, namely horizontal $x$ and vertical $y$, which can be observed separately if they are perpendicular. For the projectile motion, the projectile starts at $\vec{s}(0) = (0,0)$ with the initial velocity $\vec{v}(0) = (v_{x0},v_{y0})$ and the acceleration is constant $\vec{a} = (0, -g)$, hence

$$s_x(t) = v_{x0} \cdot t, \qquad s_y(t) = - \frac{1}{2} g \cdot t^2 + v_{y0} \cdot t \tag 1$$

where $s_x$ and $s_y$ are horizontal and vertical displacements, i.e. $\vec{s}=(s_x,s_y)$, and initial velocity in horizontal and vertical directions is:

$$v_{x0} = v_0 \cos(\alpha), \qquad v_{y0} = v_0 \sin(\alpha)$$

where $v_0$ is the initial velocity magnitude, and $\alpha$ is the angle of the projectile to the horizontal.

The displacement equations are now

$$s_x(t,\alpha) = (v_0 \cos(\alpha)) \cdot t, \qquad s_y(t,\alpha) = - \frac{1}{2} g \cdot t^2 + (v_0 \sin(\alpha)) \cdot t$$

Now you find the time at which the projectile hits the ground $s_y = 0$:

$$t_0 = 0, \qquad t_1 = \frac{2 (v_0 \sin(\alpha))}{g}$$

where $t_0 = 0$ is the obvious solution. The horizontal displacement for the time $t_1$ is:

$$s_x(t_1,\alpha) = (v_0 \cos(\alpha)) \cdot \frac{2 (v_0 \sin(\alpha))}{g} = \frac{v_0 ^2}{g} \sin(2\alpha)$$

Maximize the above equation with respect to $\alpha$:

$$\frac{d}{d\alpha} s_x(t_1,\alpha) = \frac{v_0^2}{g} 2 \cos(2\alpha) = 0$$

from which it follows $\alpha = \frac{\pi}{4}$. The maximum horizontal and vertical displacements are

$$s_x(t_1,\frac{\pi}{4}) = \frac{v_0^2}{g}, \qquad s_y(\frac{1}{2}t_1,\frac{\pi}{2}) = \frac{1}{4} \frac{v_0^2}{g}$$