Projectile Motion

To study projectile motion, we don't recommend students to recite all the equations. Since when any condition change, the whole equations changes. Furthermore, solving these kinds of problems always the knowledge of the relationship between these equations. It would be more easy to solve these kinds of problems if you deduce the equations from the basic one yourself.

Definition of Projectile Motion

Projectile motion is a part of parabolic motion. Parabolic motion has the following properties:

1. They are under constant acceleration.
2. There exist a non-zero component which is perpendicular to the acceleration so as to produce a parabolic path. If this condition is not satisfied, the motion is actually a linear motion.

Try the projectile similation java

To consider the kinematics of projectile motion, suppose an object is projected from the horizontal ground with initial velocity u at an angle to the horizontal as shown in the figure below.

Basic Equations of Projectile Motion

Consider the horizontal and vertical motion individually. Initially,

Horizontal velocity= u cos

Vertical velocity = u sin

The horizontal velocity is constant in the motion, since the acceleration is vertically downward. At time t, the velocity components are

Vertical velocity at time t = usin - gt

I think you must know that in many physics books, a lot of equations come out then. Actually those equations can be easily deduced by yourselves. Just follow our step.

In Form 5, you should know that when integrate the velocity function with respect to time, you will get the displacement equations.

So The horizontal and vertical displacements of the object are:

Horizontal displacement at time t =

Vertical displacement at time t =

To find the maximum height is just like finding out max. points of that parabolic equation. Just follow the step of finding turning point:

So from dy/dx = 0, we get u sin- gt = 0

Sub (1) into the vertical displacement equation

Max. height =

Therefore, most of the equations can be deduced by mathematical method.

So, how about other equations about projectile motion?

Equation of Trajectory

From above, we know that:

horizontal motion (x) =

Vertical motion (y) =

Eliminating t, we get

Moreover, the above equation can be transformed to a quadratic equation in tan

So, you may ask why we need 2 equations. Actually, the first one gives you the relationship between the x-y motion, you can find out the height by given the horizontal distance travelled and the projection angle.

The second equation gives you the relationship between the angle of projection and the path of travel. For example, if you want to find the projection angle in order to pass through a point or clear a obstacle, you will find this equation useful.

Let's summarized the projectile trajectory equations above:

 Form of Equations Uses x = ut cos, y = ut sin - 0.5gt2 Find out the position of object at a particular time. Find out the time taken for an object to travel to a particular position Find out the y-position by given other condition.(no time is needed) Find out the x-position by solving this quadratic equation. Find out the angle of projection. Determine the minumum velocity of projection to clear an obstacle.

Horizontal Range

Try to consider this problem like those in geometry, then it will be an easy question. What are the characteristic of Horiaontal Range? It is the horizontal distance travelled when the object hit the ground again. It is the end of the projectile motion.

So, we set y = ut sin - 0.5gt2 = 0 to find out the time of flight

Time of Flight (T) = 2u sin / g

So Horizontal Range (R) = uT cos

The above is just some examples on how questions can be set. In fact, there are far more to be tested. But we must say solving these equations is just like solving mathematical problems.