Trajectory Formula

A trajectory is the path taken up by a moving object that is following through the space as a function of time. Mathematically, a trajectory is described as a position of an object over a particular time. A much simplified example would by a ball or rock thrown upwards, the path taken by the stone is determined by the gravitational forces and resistance of air.

Some more common examples of trajectory motion would be a bullet fired from gun, an athlete throwing a javelin, satellite orbiting around the earth etc.

projectile motion

Trajectory formula is given by

 

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

Where,
y is the horizontal component,
x is the vertical component,
g= gravity value,
v= initial velocity,
$\theta$ = angle of inclination of the initial velocity from horizontal axis,

Trajectory related equations are:

\[\large Time\;of\;Flight: t=\frac{2v_{0}\,sin\,\theta}{g}\]

\[\large Maximum\;height\;reached: H=\frac{_{0}^{2}\,sin^{2}\,\theta}{2g}\]

\[\large Horizontal\;Range: R=\frac{V_{0}^{2}\,sin\,2\,\theta}{g}\]

Where,
Vo is the initial Velocity,
sin $\theta$ is the y-axis vertical component,
cos $\theta$ is the x-axis horizontal component.

Solved examples

Question: Marshall throws a ball at an angle of 60o. If it moves at the rate of 6m/s and Steve catches it after 4s. Calculate the vertical distance covered by it.

Solution:

Given,
$\theta = 60^{\circ}$
$Initial\;velocity=v_{0} = 6m/sec$
time = 4 sec

The horizontal distance is given by:

$x=V_{x0}$
$t=6\;m/sec\times 4\;sec$
$x=24\;m/sec^{2}$

$y=x\,tan\,\theta -\frac{gx^{2}}{2v^{2}\,cos^{2}\,\theta}$

$=24\,m/sec^{2}\;tan\;60^{\circ}$

$=\frac{9.8\,m/sec^{2}(15/sec^{2})^{2}}{2(6m/sec)^{2}cos^{2}\,60^{\circ}}$

$=24.5\,m/s^{2}$


Practise This Question

In the following figure, if the chords AB and CD are equal to 6cm then find the distance of the chords from the centre of circle of radius 5 cm.