# Undergraduate Mathematics/Projectile motion

Jump to: navigation, search

Template:Wikify

Parabolic water trajectory
Initial velocity of parabolic throwing
Components of initial velocity of parabolic throwing

Projectile motion is a form of motion where a particle (called a projectile) is thrown obliquely near the earth's surface, it moves along a curved path under the action of gravity. The path followed by a projectile motion is called its trajectory. Projectile motion only occurs when there is one force applied at the beginning of the trajectory after which there is no interference apart from gravity.

## The initial velocity

If the projectile is launched with an initial velocity v0, then it can be written as

${\displaystyle \mathbf {v} _{0}=v_{0x}\mathbf {i} +v_{0y}\mathbf {j} }$.

The components v0x and v0y can be found if the angle, α is known:

${\displaystyle v_{0x}=v_{0}\cos \theta }$,
${\displaystyle v_{0y}=v_{0}\sin \theta }$.

If the projectile's range, launch angle, and drop height are known, launch velocity can be found by

${\displaystyle V_{0}={\sqrt {{R^{2}g} \over {R\sin 2\theta +2h\cos ^{2}\theta }}}}$.

## Kinematic quantities of projectile motion

In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other.

### Acceleration

Since there is no acceleration in the horizontal direction velocity in horizontal direction is constant which is equal to ucosα. The vertical motion of the projectile is the motion of a particle during its free fall. Here the acceleration is constant, equal to g.[1] The components of the acceleration:

${\displaystyle a_{x}=0}$,
${\displaystyle a_{y}=-g}$.

### Velocity

The horizontal component of the velocity remains unchanged throughout the motion. The vertical component of the velocity increases linearly, because the acceleration is constant. At any time t, the components of the velocity:

${\displaystyle v_{x}=v_{0}\cos(\theta )}$,
${\displaystyle v_{y}=v_{0}\sin(\theta )-gt}$.

The magnitude of the velocity (under the Pythagorean theorem):

${\displaystyle v={\sqrt {v_{x}^{2}+v_{y}^{2}\ }}}$.

### Displacement

Displacement and coordinates of parabolic throwing

At any time t, the projectile's horizontal and vertical displacement:

${\displaystyle x=v_{0}t\cos(\theta )}$,
${\displaystyle y=v_{0}t\sin(\theta )-{\frac {1}{2}}gt^{2}}$.

The magnitude of the displacement:

${\displaystyle \Delta r={\sqrt {x^{2}+y^{2}\ }}}$.

## Parabolic trajectory

Consider the equations,

${\displaystyle x=v_{0}t\cos(\theta )}$,
${\displaystyle y=v_{0}t\sin(\theta )-{\frac {1}{2}}gt^{2}}$.

If we eliminate t between these two equations we will obtain the following:

${\displaystyle y=\tan(\theta )\cdot x-{\frac {g}{2v_{0}^{2}\cos ^{2}\theta }}\cdot x^{2}}$,

This equation is the equation of. Since g, α, and v0 are constants, the above equation is of the form

${\displaystyle y=ax+bx^{2}}$,

in which a and b are constants. This is the equation of a parabola, so the path is parabolic. The axis of the parabola is vertical.

## The maximum height of projectile

Maximum height of projectile

The highest height which the object will reach is known as the peak of the object's motion. The increase of the height will last, until vy = 0.

${\displaystyle 0=v_{0}\sin(\theta )-gt_{h}}$,

Time to reach the maximum height:

${\displaystyle t_{h}={v_{0}\sin(\theta ) \over g}}$.

From the vertical displacement the maximum height of projectile:

${\displaystyle h=v_{0}t_{h}\sin(\theta )-{\frac {1}{2}}gt_{h}^{2}}$,

so

${\displaystyle h={v_{0}^{2}\sin ^{2}\theta \over {2g}}}$ .

## Additional Equation

For the relation between the distance traveled, the maximum height and angle of launch, the equation below has been developed.

${\displaystyle h={d\tan \theta \over {4}}}$

## Notes

1. The g is the acceleration due to gravity. (9.81 m/s2 near the surface of the Earth).