A jet issuing at a velocity of 20 m/s is directed at 30° to the horizontal. Calculate the height cleared by the jet at 25m from the discharge location? Also determine the maximum height the jet will clear and the corresponding horizontal location.
It would have crossed this height also at 10.43 m from the starting point (check using equations derived in Problem 6.17).
Problem 6.17
A liquid jet at a velocity V0 is projected at angle θ. Describe the path of the free jet. Also calculate the maximum height and the horizontal distance travelled.
The horizontal component of the velocity of jet is Vxo
= V o cos θ. The vertical component Vzo
= Vo
sin θ.
In the vertical direction, distance travelled, Z, during time t, (using the second law of Newton)
Maximum horizontal reach is at θ = 45° or 2θ = 90° and for this angle it will reach half the vertical height.
This describes an inverted parabola as shown in Fig. P.6.17
Bernoulli equation shows that Zt
+ Vt
2/2g = constant along the rejectory. Vt
is the velocity at that location when air drag is neglected. Pressure is assumed to be uniform all over the trejectory as it is exposed to atmosphere all along its travel. Hence