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# Projectile motion formula

In this article find Projectile motion formula for an object fired at an angle and for the object fired horizontally. Find the relevant formula with examples for better understanding.
Projectile motion is a form of motion in which an object or particle (called a projectile) is thrown with some initial velocity near the earth’s surface, and it moves along a curved path under the action of gravity alone.
It is an example of a two-dimensional motion with constant acceleration.

A body moves in a projectile motion when it is thrown at an angle from the horizontal surface or when an object is fired or thrown horizontally from the cliff. This is also observed if an object is thrown at an angle with horizontal from cliff

## Projectile Motion of an object fired Horizontally from the top of the tower

Initial Velocity

Horizontal Component
${v_{0x}} = {v_0}$

Vertical Component
${v_{0y}}=0$

Horizontal Acceleration
$a_x=0$

Vertical Acceleration
$a_y=-g$

Equation of Motion

X-coordinate at any point
$x={v_0} t$

Horizontal Component at any point
$v_x={v_0}$

Y-coordinate at any point
$y=H – \frac {1}{2} gt^2$

Vertical component at any point
$v_y=-gt$

Trajectory Equation
$y=H-\frac {gx^2}{2v_0}$
The trajectory is a parabola

Time of Flight
$t= \sqrt {\frac {2H}{g}}$

Horizontal Range of the Projectile
$R= v_0 \sqrt {\frac {2H}{g}}$

## Projectile Motion of an object fired at an angle $\theta _0$ to horizontal

Initial Velocity

Horizontal Component
${v_{0x}} = {v_0}cos{\theta _0}$

Vertical Component
${v_{0y}} = {v_0}sin{\theta _0}$

Horizontal acceleration
$a_x=0$

Vertical Acceleration
$a_y=-g$

Equation of Motion

X-coordinate at any point
$x=(v_0cos{\theta _0})t$

Y-coordinate at any point
$y=(v_0sin{\theta _0})t- \frac {1}{2}gt^2$

Horizontal Component at any point
$v_x=v_0cos{\theta _0}$

Vertical component at any point
$v_y={v_0}sin{\theta _0}-gt$

Trajectory Equation

$y=(tan{\theta _0})x-[\frac {g}{2(v_0cos{\theta _0})^2}]x^2$
This is the equation of the parabola

Time of flight of projectile
$T=\frac {2v_0 sin{\theta _0}}{g}$

Maximum Height reached by Projectile
$H=\frac {v_0^2 sin^2 {\theta _0}}{2g}$

Horizontal Range of the Projectile

$R=\frac {v_0^2 sin {2 {\theta _0}}}{g}$

Special Cases
a. Maximum Range for a given initial velocity is obtained when $\theta _0=45^0$.
Maximum Range is given as
$R_m=\frac {v_0^2}{g}$

Example : An ball is kicked at the velocity of 18 m/s at an angle 30° from the horizontal. Find the time of flight of the ball,range of the ball. Take $g=10m/s^2$
Solution
Here $v_0=18 m/s$
$\theta _0=30$
$T=\frac {2v_0 sin{\theta _0}}{g}$
$T= \frac {2 \times 18 \times .5}{10} =1.8 sec$
$R=\frac {v_0^2 sin {2 {\theta _0}}}{g}$
$R= \frac {18^2 \times \sqrt {3}}{2 \times 10}=28.05 m$

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