A rectangular channel 6 m wide with a depth of flow of 3 m has a mean velocity of 1.5 m/s. The channel undergoes a smooth, gradu
al contraction to a width of 4.5 m.
(a) Calculate the depth and velocity in the contracted section.
(b) Calculate the net fluid force on the walls and floor of the contraction in the flow direction.
In each case, identify any assumptions made.
(a) Calculate the depth and velocity in the contracted section.
(b) Calculate the net fluid force on the walls and floor of the contraction in the flow direction.
In each case, identify any assumptions made.
1 answer:
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Answer:
Explanation:
<u>Projectile Motion</u>
In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:
Where is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.
The horizontal and vertical distances are, respectively:
The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find
Using this time in the horizontal distance, we find the Range or maximum horizontal distance:
Let's solve for
This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:
Or equivalently:
Given Vo=37 m/s and R=70 m
And