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?


2 Answers 2


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 $\alpha=45^\circ$.


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.

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

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

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    $\begingroup$ Valiant effort. But I suspect this will be decried as "a mathematical explanation". $\endgroup$ Dec 9, 2021 at 13:33
  • $\begingroup$ Possibly, although it does not use any complex calculus except maybe to derive expression for the displacement with constant acceleration which is a well-known equation. I see others tried to answer with "no calculus to prove this", but the only thing that is different is that they based their calculation on velocity instead of the displacement which I find more intuitive. Sure, I could have skipped the angle maximization line, but I used it to provide a better understanding how things are maximized in general. If you know some completely different approach I would be happy to see it. $\endgroup$ Dec 9, 2021 at 15:02

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