Fluid Flow and Projectile Motion

[dropgigawatt]

Homework Statement

Water travels through a pipe with a radius of 1 inch at a speed of 2 m/s. There is a nozzle at the end of the pipe with a radius of 1/4 inch. The pipe is horizontal and 2m above the ground. How far does the water travel in the air before hitting the ground?

r₁ = 1 in.
r₂ = 1/4 in.
V₁ = 2 m/s
y = 2 m

Homework Equations

A₁V₁=A₂V₂
A = ∏r²
Δx = V_xt

The Attempt at a Solution

I found that A₁ = 3.14 in.² and A₂ = .20 in.²
I plugged those in and found V₂ = 31.4 m/s².

Now I think I have to find t so I can work with Δx = V_xt but I'm not sure how to go about that or if that's even the direction I should be taking.

EDIT: I tried using kinematics, re-arranging V² = V₀^2 + 2aΔx. I plugged my values into -V₀^2 / 2a = Δx and got -31.4² / 2(9.81) = 50.3 m
Would that be correct?

 

[voko]

I found that A₁ = 3.14 in.² and A₂ = .20 in.²

You did not have to. All you need is the ratio $\frac {A_1} {A_2} = \frac { \pi r_1^2 } {\pi r_2^2 } = \left( \frac {r_1} {r_2} \right)^2 = 16$, so the velocity at the nozzle is 32 m/s exactly.

I plugged those in and found V₂ = 31.4 m/s².

The result is inaccurate, and the dimension is wrong.

Now I think I have to find t so I can work with Δx = V_xt but I'm not sure how to go about that or if that's even the direction I should be taking.

The rest of the problem is simple projectile motion. You know the initial velocity and the initial height. Find where it strikes the ground.

 

[Chestermiller]

The water is traveling both horizontally and vertically. They want you to find how far it travels horizontally when it hits the ground. How long does it take for an object dropped from a height of 2 m to reach the ground? How far does an object travel horizontally in that amount of time if its horizontal speed is 32 m/s?

 

[dropgigawatt]

I got it.
t = .64 s
displacement = 32(.64) = 20.4 m

Thanks guys!

 

[Nickie]

Your approach using the equation V2 = V0^2 + 2aΔx is correct. However, you need to use the acceleration due to gravity, which is approximately 9.8 m/s^2. Also, your units for V2 should be m/s, not m/s^2. Therefore, the correct calculation would be: V2 = √(V1^2 + 2gΔx) = √(2^2 + 2(9.8)(2)) = 6.26 m/s.

To find the time, you can use the equation Δx = V0t + 1/2at^2, where V0 is the initial velocity (in this case, 6.26 m/s) and a is the acceleration due to gravity (9.8 m/s^2). Solving for t, you will get t = 0.63 seconds.

Finally, to find the distance traveled in the air, you can use the equation Δx = Vxt, where Vx is the horizontal velocity, which is equal to V2 (since there is no acceleration in the horizontal direction). Therefore, Δx = (6.26 m/s)(0.63 s) = 3.94 meters.

Therefore, the water will travel approximately 3.94 meters in the air before hitting the ground.