### Introduction

Important definitions

$\to$ Projectile: A body thrown into space and allowed to move only under the action of gravity is called projectile.

$\to$ Projectile Motion: The motion shown by projectile is called projectile motion.

$\to$ Angle of Projection: The angle with the horizontal at which the body is projected is called the angle of projection.

$\to$ Velocity of Projection: The velocity with which the body is thrown is called the velocity of projection.

$\to$ Point of Projection: The point from which the body is projected in the air is called a point of projection.

$\to$ Trajectory of Projectile: The path followed by a projectile in the air is called the trajectory of the projectile.

$\to$ Range of Projection: The horizontal distance travelled by the body performing projectile motion is called the range of the projectile.

Assumptions of Projectile Motion

$\to$ The free-fall acceleration is constant over the range of motion and it is directed downward.

$\to$It is reasonable as long as the range is small compared to the radius of the Earth.

$\to$ The effect of air friction is negligible.

The projectile motion near the surface of the earth (gravity is constant) consists of two independent motions.$\\$

$\to$A horizontal motion at a constant speed.$\\$

$\to$A vertical motion subjected to a constant acceleration due to gravity.$\\$

$\textbf {Note:}$ In projectile motion acceleration is constant but velocity is changing in magnitude and direction at every instant.

### Oblique Projection

When a body is thrown into space at an angle θ with horizontal with initial velocity u from the ground. This is called oblique projection.

Few Deductions:

$\to$ In X axis initial velocity, $u_x = u \cos\theta$ acceleration, $a_x = 0$ i.e the motion is uniform.

$\to$ In Y axis initial velocity, $u_y=u \sin \theta$ acceleration $a_y=- g$ i.e the motion is uniformly accelerated.

$\to$ Horizontal distance travelled $(x)=u_x t + \dfrac{1}{2} a_xt^2= u t \cos \theta$ (Where $u_x=u \cos \theta$)

$\to$ Vertical distance covered $(y)= u_yt + \dfrac{1}{2} a_yt^2 = u t \sin \theta -\dfrac{1}{2}gt^2$

$\to$ Instantaneous Velocity (v):

Magnitude: $\sqrt{v_x^2 + v_y^2}$

Direction: $\tan \alpha = \dfrac{v_y}{v_x}$

$\to$ Equation of trajectory: $y=x \tan θ - \dfrac{gx^2}{2u^2cos^2\theta}$. This is the equation of trajectory in projectile motion

$\to$ $y \propto x^2$ i.e. The projectile motion is always parabolic in nature.

Important Terms in Projectile:

$\to$ Time of flight (T) = $\dfrac {2u_y}{g}=\dfrac{2u \sin \theta}{g} = \sqrt{\dfrac{2H_{max}}{g}} = \dfrac{R}{u \cos \theta}$

$\to$ Horizontal Range(R)=$u_x T = \dfrac{2u_xu_y}{g}=\dfrac{u^2 \sin 2 \theta }{g}=4H_{\text{max}} \cot \theta$

$\to$ Maximum Height ($H_{max})=\dfrac{{u_y}^2}{2g} = \dfrac{u^2\sin^2\theta}{2g} = \dfrac{gT^2}{8} = \dfrac{1}{4} R \tan \theta$

For a projectile to obtain same range the angles of projection should be $\theta$ and $90- \theta$. (In a condition that speed is constant).

For maximum horizontal range, the angle of projection should be 45 degrees.

In projectile motion, the horizontal component of velocity $(u \cos \theta)$, acceleration (g) and mechanical energy remains constant while speed, velocity, the vertical component of velocity $(u \sin \theta)$, momentum kinetic energy and potential energy all change. Velocity and KE are maximum at the point of projection while minimum (but not zero) at the highest point.

Angle between acceleration and velocity gradually decreases from $90+\theta$ to $90- \theta$ while acceleration and velocity are right angles at the highest point.

At the highest point,$\\$

$\to$ Velocity(V) $= u\cos\theta \\$

$\to$ Momentum$=mu\cos\theta = p\cos\theta \\$

$\to$ Kinetic Energy$=E\cos^2\theta$

(Substitute $v=u\cos\theta$ in general eqn of K. E)$\\$

$\to$ Potential Energy$=E\sin^2 \theta$

From point of projection to maximum height$\\$

$\to$ Change in direction $=\theta$$\\$

$\to$ Change in velocity $= u\sin\theta \\$

$\to$ Change in momentum $= mu\sin \theta \\$

$\to$ Decrease in K.E(increase in P.E) $= E\sin^2\theta \\$

$\to$ Change in speed $= u-ucos\theta =2u\sin^2\dfrac{\theta }{2} \\$