A fire hose held near the ground shoots water at a speed of 7.0 m/s.?

Question

At what 2 angles could the nozzle point in order that the water would land 2.0 m away

Answer 1

Let's call the angle with the horizontal x. Then the vertical component of the water's initial velocity is 7 sin x and the horizontal component, which is constant, is 7 cos x metres/s. It will take the water (7/10) sin x seconds to reach max. height (gravitational acceleration=10m/s^2), and the same time to reach the ground again, so it will be in the air 1.4 sin x secs. During this time it is moving horizontally at 7 cos x m/s so it will land 7times 1.4 sinx cos x metres away, that is 9.8 sin x cos x metres. For this distance to equal 2.0 metres, sinx cos x =2/9.8=0.204. So just find the two solutions of this equation between 0 and 90 degrees. I'd do it graphically.

Answer 2

The independent variable is the angle of the nozzle (call it A).
We need to find 2 values of A for which the water will land 2m from the nozzle.

We know the velocity of water leaving the nozzle is 7m/s. If we choose a value for A, we can determine the horizontal and vertical components of that velocity (7cosA and 7sinA, respectively).

From the vertical component, we can figure out how long the water will be in the air before it hits the ground: With gravity slowing its vertical velocity by 9.8m/s each second, it will decrease to a vertical velocity of 0 in 7sinA/9.8 seconds. At that point it will be at its highest point and will start falling. It will then take the same amount of time to return to the ground. Total time: 2(7sinA/9.8).

From the horizontal component and the total time before hitting the ground, we can calculate how far the water will travel before hitting the ground. It will travel (horizontally) at 7cosA m/s for a period of 2(7sinA/9.8) sec, for a total distance of:
(7cosA)2(7sinA/9.8) = 98sinAcosA/9.8 m = 10sinAcosA.

We simply need to find the value(s) of A for which this calculated distance equals 2 meters. Fortunately, we can simplify the expression by recognizing that 2sinAcosA = sin(2A).
So the total distance = 5sin(2A).
So we have 5sin(2A) = 2, or sin(2A) = .4

All you have to do is determine the two angles for which the sin is .4.
Then divide these angles by 2 to find the values of A for which the water travels 2 m horizontally.

Note: zee prime's answer is correct, but 2 comments:
1. He rounds the metric value of g to 10m/s^2. I used the value of 9.8, which turns out to work out nicely with the values in this problem (you get 98/9.8 = 10).
2. He doesn't give you a way to solve for the angle. The "double angle" formula I gave you allows you to find the angle directly, instead of "graphically," as zee prime suggests.

Good luck!

Answer 3

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