what is the derivation of motion of trajectory and its explanation
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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.
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
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.
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
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