CodeCogs

The flight of a Frictionless Projectiles

Interface

#include <codecogs/engineering/projectiles/frictionless.h>

using namespace Engineering::Projectiles;

double height(double R, double theta, double t, double h=0)
double max_height(double R, double theta)
double impact_velocity(double R, double theta, double h)
double position(double R, double theta, double x, double h=0)

Overview

With a Frictionless system the influence of air on the projectile is ignored. On earth this leads to a optimistic estimate for the distance travelled by the projective, although on an airless planet/moon these calculations would be appropriate.

In this module, the initial velocity of the projectile is R, released at an angle $\theta$ at a height h. i.e.

The only force acting on a frictionless projectile is that of gravity. As can be seen from the diagram the initial launch velocity,R, is resolved into it's vertical and horizontal components. As there are no forces in the horizontal direction, the horizontal component of the initial velocity lasts for the duration of the flight. Therefore the main equations of motion need only be applied in the vertical direction:

$s=ut -{ \frac{1}{2}}{g t^2}$

(1)

$v^2=u^2 + 2as$

(2)

$v=u+at$

(3)

Where

u is the initial vertical velocity [m/s]
v is the final vertical velocity [m/s]
s is the vertical distance traveled [m]
t is the time taken [s]
a is acceleration [m/s\^2]

FUNCTION

height

In the vertical direction the only force acting on the projectile is Gravity, g, which is assumed to be constant. With an initial vertical velocity of $u= R \sin \theta$ , then Eq (1) gives

$s(t)=R \sin\theta \cdot t -{ \frac{1}{2}}{g t^2}$

(4)

which is the vertical height of the projectile at any given time.

Parameters

is the initial velocity of the projectile [m/s]

theta

is the angle the projectile is released from the horizontal [degrees]

the time from when the projectile is launched

Default value = 0

Interactive Calculator

theta

Result –

FUNCTION

max_height

At its greatest height, the vertical velocity is zero, therefore from Eq (2) the maximum vertical height is:

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

(5)

We can show from Eq(3) that the time to reach this maximum is

$t=\frac{R\sin\theta}{g}$

(6)

Parameters

is the initial velocity of the projectile [m/s]

theta

is the angle the projectile is released from the horizontal [degrees]

Interactive Calculator

theta

Result –

FUNCTION

impact_velocity

The velocity of the projectile as it hits the ground is

$V=\sqrt{v_{horz}^2 + v_{vertl}^2}$

(7)

where

v_horz = $R \cos \theta$
v_vert is computed from Eq (2), with the initial assumptions that u = 0 (at maximum height) and height of the projectile from its final landing point is

$s=h+\frac{R^2 sin^2 \theta}{2 g}$

(8)

Therefore

$v_{vert}^2 = 2g(\frac{h+R^2\sin^2\theta}{2g})$

(9)

$v_{vert}=\sqrt{h+R^2\sin^2\theta}$

(10)

Parameters

is the initial velocity of the projectile [m/s]

theta

is the angle the projectile is released from the horizontal [degrees]

the initial vertical height of the projectile

Interactive Calculator

theta

Result –

FUNCTION

position

To conveniently plot the position of the projectile, it is necessary to compute the height of the projectile, y, for a given horizontal position x.

Therefore at a given x position, the time taken to reach that point is merely:

$t=\frac{x}{R\cos\theta}$

(11)

From Eq(3) and maximum flight time of the projectile is:

$t_{max}=\frac{1}{g}\left [\sqrt{h+R^2\sin^2\theta} + R \sin\theta \right ]$

(12)

which serves as an upper limit for Eq(11)

Using the height function (above), y is computed.

Parameters

is the initial velocity of the projectile [m/s]

theta

is the angle the projectile is released from the horizontal [degrees]

the horizontal position of the projectile.

the initial vertical height of the projectile.

Interactive Calculator

theta

Result –