## Understanding the Projectile Motion

In kinematics, we study the motion of various objects and the path taken by the object while changing its position by the effect of a force applied to it. the path can be regular and quantifiable like a straight line or can be irregular without any definite direction. The other parameters such as speed and acceleration are also measured by observing the different values of the object while in motion. We also have studied the attraction force that the earth exerts on every object over its surface. It is known as the force of gravity and is represented by the symbol 'g'. Thus, any object moving near the surface of the earth is influenced by this force and gets accelerated downward towards the ground. the object in such a condition is said to be in a projectile motion.

The path that it attains is known as the trajectory. The best example of a projectile motion is when you throw a stone into the air. It travels to a certain height by the force provided by you and then falls freely under the influence of gravity. But the horizontal flight of the object remains unchanged and we can observe a curved path of motion. For the same speed, it travels the maximum distance when we throw a stone at 45 degrees angle to the ground. A batsman in the game of cricket also tries to hit the ball 45 degrees to get the ball away from the boundary. The same principle is used in various other areas of application. Ballistic missiles that we hear about all the time in the news also use this principle to travel the maximum distance possible during wars.

The equations to find out the maximum distance, maximum height and time of flight are mentioned in the chapters of motion.

### Trajectory Equation

A trajectory or flight path in physics is defined as the path that an object in motion having some mass follows through space as a function of time. Hamiltonian mechanics via canonical coordinates are used to define the trajectory of an object in classical mechanics. Hence both position and momentum simultaneously define a complete trajectory. For example, the motion of a ball or a big rock was thrown upwards or the motion of a satellite or a bullet fired from a gun.

### Derivation of Equation of Trajectory

The equation of trajectory derivation is as follows ---

\[ y = x~tan~\theta - \frac{gx^{2}}{2 v^{2} cos^{2} \theta}\]

where, ‘x’ is the horizontal component of the trajectory.

‘y’ is the vertical component of the trajectory.

‘g’ is the gravitational force of attraction.

‘v’ is the initial velocity of the object.

‘ϴ’ is the angle of elevation of the trajectory.

### Some Important Trajectory Related Equations are as Follows

Time of Flight,\[T = \frac {2v_{0} sin\theta} {g}\]

Maximum Height Reached, \[H = \frac{v{_{0}}^{2} sin^{2} \Theta}{2g}\]

Horizontal Range , \[R = \frac{v{_{0}}^{2} sin2 \Theta}{g}\]

Where ‘\[v_{0}\]’ is the initial velocity

‘sinϴ’ is the vertical component of the y-axis

‘cosϴ’ is the horizontal component of the x-axis.

Thus the trajectory equation along with some important formulae has been derived. Objects will follow a vertically symmetric path when projected from and land on the same horizontal surface. The flight time depends on the initial velocity of the projectile and the angle of projection. The maximum height of the projectile is reached when the velocity of the object is zero. The range of the projectile depends on the initial velocity of the object.

This article explains the trajectory formula and the derivation of the equation of trajectory. Some FAQs have been added for a better understanding of the topic for the students.

**1. What are all the equations of the trajectory?**

In the trajectory path equations are derived for the maximum height reached above the ground, the horizontal range of travel achieved by the body and the total time of flight or otherwise known as the total time it took before falling onto the ground. to derive the equations both the horizontal component ' x ' and vertical component ' y ' are taken into account. The vertical component is obtained by the addition of the gravity force of attraction and the vertical velocity of the object. While the horizontal component is only the horizontal velocity.

2. Do we have to remember all the formulas of this topic?

The topic of projectile motion is a very important concept to understand in the chapter on Motion in the plane of class 11. Questions are asked from this chapter frequently which are based on the equations of the trajectory produced by a rigid body while travelling in a projectile motion. To find out the answers to the questions based on these concepts students have to use the equations for that particular problem. So it would be actually easy if you remember the equations and apply them during term exams or competitive exams.

**3. What is the difference between speed and acceleration?**

When a body moves from one position to another then it can be either uniform motion or non-uniform motion. When an object moves in uniform then the distance covered is equal in every unit of time. But if it is non-uniform motion then the distance covered per unit of time is not equal. The distance travelled per unit of time is known as the speed of the object. It is often assumed uniform motion. when in non-uniform motion the speed increases or decreases constantly. The measure of the increase and decrease of speed per unit time is known as acceleration and deceleration respectively.

**4. Why don't aero planes undergo trajectory motion while flying in the sky?**

It is quite common to see the objects in a projectile motion travelling a path of trajectory. These objects in this context start with the application of force applied at the point of origin which carries it forward. And because of the influence of gravity, it comes down to the ground and ends the trajectory. But aeroplanes are machines having engines that propel them constantly. And the aerodynamics of the aeroplanes keep it lifted in the air while it moves and prevents it from coming down and making any trajectory.

**5. Prove that a gun will shoot three times as high when its angle of elevation is as when it is but covers the same horizontal range.**

Let ‘ϴ’ be the angle of elevation. When ϴ=60^{o}, then maximum height,

\[H_1 = \frac{u^2 sin^2 60^0}{2g} = \frac{3u^2}{8g} \]

where ‘u’ is the initial velocity of the bullet, ‘g’ is the gravitational force of attraction. Horizontal Range,

\[R_1 = \frac{u^2 sin(2 x 60^0)}{g} = \frac{\sqrt{3u^2}}{2g} \]

When ϴ=30^{o}, then maximum height,

\[H_2 = \frac{u^2 sin^2 30^0}{2g} = \frac{u^2}{8g} \]

Horizontal Range,

\[R_2 = \frac{u^2 sin(2 x 30^0)}{g} = \frac{\sqrt{3u^2}}{2g} \]

Hence, \[H_1 = 3H_2 \] and \[R_1 = R_2 \]

Therefore a gun will shoot three times as high when its angle of elevation is as when it is but covers the same horizontal range.

**6. What are the applications and give two examples of the projectile trajectory equation?**

A projectile is any object that is cast, fired, flung, heaved, hurled, pitched, tossed, or thrown. Some applications of the projectile trajectory equation in real life are as follows:-

Playing sports like basketball and football as a curve is generated when the ball is thrown or kicked.

Space Explorations require the help of the trajectory formula.

During situations of war and flood operations also the trajectory formula is essential.

**Examples of the projectile trajectory **

A bomb is dropped from a plane during the war.

An athlete throwing a javelin.

A ball is kicked or thrown as in football, basketball and cricket respectively.

**7. What is the change in momentum between the initial and final points of the trajectory path, if the range is maximum?**

If we know that the range is maximum. The angle of projection, ϴ=45^{o}. Now, since the initial velocity horizontally remains the same throughout the horizontal part, there is no change in momentum in the horizontal. The magnitude of vertical velocity = u sin 45^{o}, where ‘u’ is the initial velocity of the object. The vertical velocity will be equal in magnitude but opposite in direction at the required points. Hence, the change in momentum = 2mu * sin 45^{o} = √2mu , where ‘m’= mass of the object. So the change in momentum will be the product of the square of two and mass and initial velocity.