How Does a Gardner Shower Jet's Range Impact the Wall's Wetted Area?

[succhi]

Homework Statement

a gardner shower jet is placed at a distance 'd' from the wall of a building . If 'R' is thr maximum range of the jet taht is prduced when the bowl is connected to the nose of a nose of a fire engine show that the portion of the wall that is hit by jet of water is bounded by a parabola whose height is (Rd-d^2) /R and the breadth is underroot (R^2-d^2)
ans it how o do it ?

Homework Equations

R=u^2(sin2a)/g

The Attempt at a Solution

tried to find range but the distance between the wall and the spinter does matter

 

[nvn]

succhi: There appears to be a typographic mistake in the problem statement. The breadth should be 2*(R^2 - d^2), not (R^2 - d^2). Please correct me if I am misinterpreting. Also, even though the problem statement says the water nozzle is on a fire engine hose, the problem contradictorily still assumes the nozzle is at ground level. Therefore, assume the nozzle is at ground level.

In item 3, you said you tried to find the range. This does not make sense, because the range is already given, and is R. And you said, "... but the distance between the wall and nozzle does matter," as if to imply that the wall distance is unknown. This does not seem to make sense either, because the wall distance is already given, and is d. The relevant equation you listed does not appear to be relevant. What is u, and what is "a" in that equation?

Relevant equations for this problem would be the basic kinematics equations for uniform motion and uniformly-accelerated motion. These equations will be listed in any physics or dynamics textbook.

The maximum water stream range, R, occurs when the angle between the nozzle and the ground is theta = 45 deg, right? The initial velocity of the water stream exiting the nozzle, vo, is unknown. The horizontal velocity of the water stream is vx = vo*cos(theta). The flight time of the water stream when it hits the ground at x = R is t3 = R/vx = R/[vo*cos(theta)]. The initial vertical velocity of the water stream exiting the nozzle is vyo = vo*sin(theta). Substitute the expression for t3 into one of the kinematics equations involving time, then solve for vo.

Hopefully the above will give you enough information to get started. Now continue using the kinematics equations to solve the problem.

 

[thomate1]

I would approach this problem by first analyzing the given information and identifying the relevant equations and variables. From the given information, we know that a Gardner shower jet is placed at a distance 'd' from the wall of a building, and the maximum range of the jet is 'R' when connected to a fire engine. We also know that the equation for range is R=u^2(sin2a)/g, where u is the initial velocity, a is the angle of projection, and g is the gravitational acceleration.

To solve for the portion of the wall that is hit by the jet of water, we need to consider the trajectory of the water from the jet. Since the water is projected at an angle, it will follow a parabolic path. The height of the parabola can be determined by the equation h=Rd-d^2/R, where h is the height and R is the maximum range.

To find the breadth of the parabola, we can use the Pythagorean theorem. Since the jet is placed at a distance 'd' from the wall and the maximum range is 'R', the breadth can be calculated as the square root of (R^2-d^2).

Therefore, the portion of the wall that is hit by the jet of water can be described by a parabola with a height of (Rd-d^2)/R and a breadth of √(R^2-d^2). To visualize this, we can plot this parabola on a graph with the x-axis representing the distance from the wall and the y-axis representing the height of the parabola.

I hope this helps to explain the solution to this problem. If you have any further questions or require clarification, please let me know.