Physics Applications: Projectile Motion

Projectile Motion

Unlike the previous two physics explanations, projectile motion deals with how far an object will travel through the air to reach its destination. There are numerous equations used to solve projectile motion problems, but the two equations which are most important in our case are the ones which involve the displacement of x and y.

δy = (Viy)t + .5at²

where δy is the displacement of the projectile's elevation,

Viy is the initial y velocity of the projectile's elevation,

t is time in seconds,

and a is the acceleration due to gravity

Basically, our first equation solves for how far the projectile will fall to the ground from the moment it's released. Using t, the function accumulates the projectile's velocity (hence the Viyt) and weight (hence the .5at²) onto the resulting displacement; one can thus think of it as an evaluated integral.

δx = (Vix)t

where δx is the projectile's horizontal displacement when it collides into the ground,

Vix is the initial x velocity of the projectile,

and t is the time in seconds

This next equation is far less complicated thanks to the fact that x velocities do not generally change for airborne objects. That way, all we need to remember is that velocity is how far something travels in a certain amount of time.

In a perfect world, the farthest launch angle for any projectile is usually about 45 degrees. Due to certain factors such as air resistance, that hardly ever happens except one is inside a vacuum, but because of both facts that there's not much of a difference between 43 and 45 anyways and that a 45 degree angle always means the absolute values of the x and y velocities are the same at start, we'll use a 45 degree angle as the angle we aim for.

After tossing those equations around together, we came up with an equation that calculates how fast a catapult must launch at a 45 degree angle in order to reach a certain distance, also given the vertical displacement of the projectile from launch to impact:

V = Sqrt( 9.8(δx)² / (δx-δy) )

where V is the magnitude of the 45 degree launch that must occur,

Sqrt() is a square root function,

δx is the horizontal displacement,

and δy is the vertical displacement

Now, let's plug in some values into this equation:

7.92 meters (26 feet) will substitute δx. Note that this includes both the required distance of 25 feet and the distance behind the front of the catapult which the ball launches from.

2.13 meters (7 feet) will substitute δy. Do not confuse this with the actual height of the catapult, as anything taller than five feet is unwanted.

One calculation later, we get the velocity needed to reach 25 feet: 10.3 meters per second. PLEASE NOTE, however, that we did not include air friction, so to make our catapult throw farther, we want to build the catapult so that it throws faster than that.