Projectile Motion


Quick
projectile - any body that is given an initial velocity and then follows a path determined entirely by the effects of gravitational acceleration and air resistance.


Details

Projectile motion takes into account the basic concepts:
     1. vectors
     2. sin, cos, tan, for triangles
     3. position, velocity, and acceleration
     4. effect of gravity on acceleration

If these concepts are understood then projectile motion should be easy to understand.

A ball is shot out of a cannon, at an angle to the horizon. The ball follows an arc in the air. But if that motion is split up into horizontal and vertical components, using basic concepts 1 and 2, it is relatively simple. This is because the horizontal motion is constant, it is the same. And the vertical motion is the motion due to gravity. Then the total motion is a function of the constant horizontal motion and the vertical motion due to gravity.

If a ball is shot in the air at an angle, and it is desired to know how far the ball went, well, horizontal distance equals the horizontal velocity the ball traveled at, multiplied by how long the ball was in the air. So, the time the ball was in the air should be found. To get this, the vertical velocity should be examined, because gravity determines how long the ball is going to be in the air. So to find the time, the relation for gravity should be used.

Typically, projectile motion is only in two dimensions, because a projectile moves forward horizontally with a velocity, but is affected vertically because of gravity.

Consider a particle moving in a vertical plane with an intitial velocity

u
0. The acceleration of the particle is the acceleration of gravity. As stated earlier, the horizontal motion is constant. Of course, this is assuming that air friction is negligible.

Then,


u
0 = u0x + u0y

The components u0x and u0y can then be found if we know the angle θ0 between

u
0 and the positive x direction:

u0x = u0 cos θ0 and u0y = u0y sin θ0

During its two-dimensional motion, the projectile's position vector and velocity vector is constant and always directed vertically downward. The projectile has no horizontal acceleration.

In projectile motion the horizontal motion and vertical motion are independent of each other.

This feature allows us to break up a problem involving two-dimensional motion into two separate and easier one-dimensional problems, one for the horizontal motion and one for the vertical motion.


Horizontal Motion

Because there is no acceleration in the horizontal direction, the horizontal compnent ux of the projectile's velocity remains unchanged from its initial value u0x throughout the motion. At any time t, the projectile's horizontal displacement xx0 from an initial position x0 is given by the equation:

xx0 = u0t +
1
2
at2

Since a = 0:

xx0 = u0xt

Because u0x = u0 cos θ0, this becomes:

xx0 = (u0 cos θ0)t


Vertical Motion

The vertical motion is the motion for a particle in free fall. For an object in free fall, the acceleration is constant. The following equation is used:

 yy0 = u0yt
1
2
gt2 = (u0 sin θ0)t
1
2
gt2

where the initial vertical velocity component u0y is replaced with the equivalent u0 sin θ. Similarly:

uy = u0 sin θ0gt

and

uy2 = (u0 sin θ0)2 – 2g(yy0)t


Equation of Trajectory

t can be eliminated between Eq1 and Eq2 by solving Eq1 for t and substituting into Eq2. After some rearrangement:

 y = (tan θ0)x
gx2
2(u0 cos θ0)2

If this equation is entered into a calculator, for a given angle and initial velocity, the trajectory can be seen when plotted. x0 = 0 and y0 = 0 because they are at the origin. Notice that the above equation follows the form y = ax + bx2, which is the equation for a parabola.


Horizontal Range

The horizontal range R of the projectile is the horizontal distance the projectile has traveled when it returns to its initial (launch) height. To find the range R, xx0 = R is put into Eq1 and yy0 = 0 is put into Eq2, obtaining:

R = u0xt

0 = (u0 sin θ0)t
1
2
gt2

Eliminating t yields:

R =
2u02
g
sin θ0 cos θ0

Using the identity sin 2θ0 = 2 sin θ0 cos θ0

R =
u02
g
sin 2θ0

R has its maximum value when sin 2θ0 = 1, which corresponds to 2θ0 = 90° or θ0 = 45°.