To study **Physics Projectile Motion**, we don’t recommend students to recite all the equations since when any conditions change, all the equations change. Furthermore, solving these types of problems always requires the knowledge of the relationship between the equations of Physics Projectile Motion. It would be easier to solve these kinds of problems if one deduces/ derives the equations from the basic forms of equations in Physics Projectile Motion. Engineering in Kenya has more articles.

# Definition of Physics Projectile Motion

Physics Projectile Motion is a form of parabolic motion. Parabolic motion in physics has the following properties;

- They are under a constant acceleration.
- There exists the acceleration which is perpendicular to a non-zero component so as to produce a parabolic path. If this condition is not met, then the form of motion is actually a linear motion.

To consider the kinematics of Physics Projectile Motion, suppose an object is released from the horizontal ground with initial velocity U at an angle α to the horizontal;

## Basic Equations of Physics Projectile Motion

Consider horizontal and vertical forms of motions individually. Initially,

Vertical velocity = u sin α

Horizontal velocity= u cos α

The horizontal velocity in Physics Projectile Motion is a constant in the motion, since the acceleration is vertically downward/ due to gravity. For a time t, the velocity components are

- Vertical velocity for a time t = u sin α – gt

I think one must know that in many physics books, a lot of equations come out from one given, like the one above. Actually those equations in Physics Projectile Motion can be easily deduced if one has the basics in physics. In Form 3 (for the Kenyan system), one should know that when one integrates the velocity function with respect to time, you will get the equations for displacement.

Therefore, the horizontal and vertical displacement equations of the object are obtained by integrating the horizontal and vertical equations respectively.

Find the maximum height in Physics Projectile Motion is just like finding out the maximum points of a parabolic equation. Calculus is also largely involved, in this case differentiation.

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

Substitute (1) into the vertical displacement equation to get the maximum height.

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

So, how about other equations about projectile motion?

**Equation of Trajectory in Physics Projectile Motion**

From above, we know the equations for horizontal motion and also vertical motion. Moreover, the *Physics Projectile Motion* equations can be transformed to quadratic equations using tan. So one may ask why we need two equations. Actually, the first one gives the relationship between the x and the y motions. One can find out the maximum height given the horizontal distance travelled and the projection angle in Physics Projectile Motion.

The second equation gives the relationship between the path of travel and the angle of projection (α). For example, if one wants to find the projection angle (α) in order to pass through a point or clear an obstacle, you will find the equation very useful.

Let’s summarized the projectile trajectory equations in Physics Projectile Motion above;

From of Equations;

- x = ut cos α,
- y = ut sin α – 0.5gt2

- Determine the minimum velocity of the Physics Projectile Motion to clear an obstacle.
- Find out the position of object at a certain time.
- Find out the time taken for the object to travel to a particular position.
- Find out the x-position by solving this quadratic equation.
- Find out the y-position by given other conditions (no time is needed).
- Find out the angle of projection.

### Conclusion in Physics Projectile Motion; Horizontal Range

Try and consider these problems like those in geometry, and Physics Projectile Motion will be an easy question. What are the characteristic of the horizontal range? It is the horizontal distance travelled when an object hits the ground again. It is the measure of the displacement from where the object was released to the end side of the projectile motion.

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

Time of flight (t) = 2u sin α/ g

So, the horizontal range (R) = ut cos α

The above is just some examples on how questions for Physics Projectile Motion can be set. In fact, there is far more to be tested. But we must say that solving these equations is just like solving normal mathematical problems in Physics Projectile Motion.