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Trajectory of a free jet

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In a free jet the pressure is atmospheric throughout the
trajectory.

V_{ox} = V_{o }cosq
= constant = V_{x}

*
*
V_{oy} = V_{o} sinq

*x* = V_{ox} t

*y* = V_{oy} t – gt^{2}/2

eliminating t gives,

*y* = *x* V_{oy}/V_{ox} –
g*x*^{2}/(2V_{ox}^{2})

i.e,

*y* = *x* tanq –
g*x*^{2}/(2V_{o}^{2 }cos^{2}q)

This is the equation of the trajectory.

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At the point of maximum elevation, V_{y} = 0 and
application of Bernoulli’s law between the issue point of jet and the
maximum elevation level,

V_{o}^{2}/(2g) = V_{ox}^{2}/(2g)
+ *y*_{m}

Since, V_{o}^{2}/(2g) =
V_{ox}^{2}/(2g) +
V_{oy}^{2}/(2g)

we get,

*y*_{m} = V_{oy}^{2}/(2g)

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Last Modified on: 14-Sep-2014

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