Calculating Max Height of Water Squirted from Garden Hose

AI Thread Summary
To calculate the maximum height of water squirted from a garden hose nozzle, the projectile motion equations are essential. The nozzle's radius and the velocities of water in both the hose and nozzle are crucial for determining the height. The expected answers are 42 meters for the nozzle attached and 0.4 meters without it, indicating a significant difference due to the nozzle's effect on velocity. If the calculations are incorrect, reviewing the use of projectile motion equations is necessary. Proper application of these equations will yield the correct results.
narutoish
Messages
25
Reaction score
0

Homework Statement



A nozzle with a radius of 0.250 cm is attached to a garden hose with a radius of 0.800 cm. The velocity of the water in the hose is 2.80 m/s and the velocity of the water in the nozzle is 28.7 m/s.s

(a) Calculate the maximum height to which water could be squirted if it emerges from the nozzle.

(b) Calculate the maximum height to which water could be squirted if it emerges with the nozzle removed, assuming the same flow rate.




Homework Equations



I might be missing something but don't I only have to use the projectile motion equation and solve for the delta y?

The Attempt at a Solution



I tried using the projectile motion equation but I don't get the right answer, and I don't know how to make use of the radius information.

The answers are supposed to be a) 42m and b) .4m I just need to know how it's solved
 
Physics news on Phys.org
Those answers can indeed be obtained with just the projectile motion equations. If you cannot obtain them, you must be making a mistake somewhere. Show your work.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

Similar threads

Back
Top