Will A Vertical Projectile Travel the Same Distance as A Horizontal Projectile

Question

If a projectile is fired vertically from an initial height, $h$, above ground level, and an identical projectile is fired with equal initial velocity from the same initial height in the horizontal direction, will these projectiles travel the same distance?

The distance of the vertically shot projectile is to be measured as the change in height from it's initial height to the apex of its trajectory. The distance of the horizontally shot projectile is to be measured as the horizontal distance traveled before the projectile touches the ground. Also, after the initial velocities are imparted to both projectiles, no other external force acts on either projectile, with the exception of gravity, for the duration of their flights.

Answer 1

It depends on the initial height above the ground from which both projectiles are fired. I believe you can see that if I fire the vertical projectile from ground level (that is zero distance above the ground), it will travel some distance into the air. But if I fire the horizontal projectile from ground level, the projectile will travel zero distance before touching the ground (because it was already on the ground).

So, then let's say we lift the projectile a small distance $h$ above the ground, and then repeat the experiment. The vertical projectile will travel upward, being constantly decelerated by gravity until, finally, it reaches a velocity of zero. At this point it is at the apex of it's trajectory, some distance Δh above it's starting height. The horizontal projectile will also be acted upon by gravity, and as it moves horizontally, it will also be pulled down to Earth. It will continue to travel horizontally until gravity eventually pulls it to ground level, at which point it would have traveled some horizontal distance Δx.

Now, consider these two concepts:

1) The vertical projectile will always travel the same Δh no matter what height it is fired from. Say, it starts 5ft above the ground and reaches an apex at 10ft. Then, if it starts at 10ft and the same force is imparted it will reach an apex at 15ft. Either way, the change in vertical distance is the same. It is important to note that this is only true given that the height of the projectile remains fairly close to the Earth's surface. The force of gravity does become weaker as $h$ increases, but unless we are talking about very large values of $h$, the difference is negligible for this problem.

2) The horizontal projectile will travel a different Δx depending on the initial height it is fired from. Say, again, that it starts 5ft above the ground. It will have a certain amount of flight time before it reaches the ground. Now, if it is fired from 10ft above the ground, the amount of time it has to travel horizontally before it hits the ground is greater. Greater flight time, means more horizontal distance covered: greater Δx.

Therefore, if the vertical flight distance is independent of initial height, but the horizontal flight distance is dependent on initial height, that should tell you that the two distances will only be the same for select circumstances (if they are ever the same at all). In this case there is a specific relation between the initial magnitude of velocity of the projectiles to the projectiles' initial height above the ground that will allow for the two distances to be the same. It is possible to find the exact equation for the relationship, but I won't go into it here.

In other words, yes, the two distances can be the same under select circumstances, but by all means, they are not guaranteed to be the same. If one is interested in the mathematics behind these concepts, I would suggest picking up an intro level physics textbook

Answer 2

The vertical distance traveled above $h$ in the case of a vertically launched projectile is given by the following expression:

$\Delta y = \frac{v_0^2}{2g}$

The horizontal distance $\Delta x$ travelled by a horizontally launched projectile before hitting the ground is given by:

$\Delta x = v_0 \cdot t_{\text{flight}} = \frac{v_0}{\sqrt{2gh}}$

Where $t_{\text{flight}}$ is the time it takes for the projectile to fall from height $h$ to $0$.

Here, the vertical distance is independent of initial height but the horizontal distance is not. Therefore, whether or not a projectile would travel the same distance horizontally as it would vertically depends on the initial height $h$ of the projectile and will only be the same if $h = \frac{v_0^2}{8g^3}$.