Solving Physics Problem with Garden Hose

In summary, the conversation revolved around a physics problem involving a garden hose shooting water at 6.5 m/s and traveling a distance of 2 m. The problem was to find the angle at which the hose must be for the water to reach 2 m. Several solutions were suggested, including using the equation for range of a projectile and decomposing the velocity into horizontal and vertical components. Ultimately, the formula d = -v^2*sin2@/g was used to solve the problem.
  • #1
Physik
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0
Hello. I've been fustrated all day because I couldn't figure this problem out. I know, it may be simple, but trust me, I almost ripped my hair out trying to solve this problem. I would have been able to figure this problem out last year in basic physics, but I have forgotten 75% of physics I learned last year. Any help at all would be great.

Problem: A garden hose lying on the ground shoots out water at 6.5 m/s. The water travels 2 m. Find the angle at which the hose must be for the water to reach 2 m.

http://img381.imageshack.us/img381/117/prob5va.png
 
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  • #2
try using the equation for range of a projectile

R = (v^2 * sin (2 * @)) / g
 
  • #3
well the answer is pretty simple .. its 13.8@
 
  • #4
That's no way to help him!

You just need to remember that the key to these sorts of problems is Time.

Decompose the velocity:
v(hor)=v*cos@
v(vert)=v*sin@

The horizontal velocity will never change, as there is no horizontal acceleration (assuming friction is barred). So the time the object should stay in the air is:

d=v*cos@*t,
t=d/(v*cos@)

Now how long does the object stay in the air? It has a 'parabolic' path, so it's final vertical velocity is the negative of its initial. There is a constant acceleration, gravity, so the time it spends in the air is:

a*t = v(final) - v(initial) [now, as said, the v(final) is just negative v(initial)]
g*t = - 2 v(initial) [where g = -9.81 m/s^2]
[It is important to remember that the 'v(initial) here is the vertical velocity, which was v*sin@]
g*t = -2*v*sin@
t = -2*v*sin@/g

Now we have 2 expressions for t, put them together:
d/(v*cos@) = -2*v*sin@/g
d = -v^2*(2*sin@*cos@)/g [double-angle identity: 2sin@cos@=sin2@]
d = -v^2*sin2@/g, again where g=-9.81

That's where mathmike's formula came from. Now just plug in.
 
  • #5
Well, that's no way to help him either...

When the water comes out of the hose at 6.5m/s, there is a horizontal component to its velocity and a vertical component to its velocity.

The horizontal velocity is given by: [tex] v_h=6.5cos\theta[/tex],

while the vertical velocity is given by: [tex] v_v=6.5son\theta [/tex]

Remember, the duration of the projectile in motion is determined by the vertical velocity. The horizontal velocity is always constant assuming no viscous forces act on the projectile.

the time of flight is given by:
[tex] a=\frac{v-u}{T} [/tex]
[tex] T = 2(\frac{0-6.5cos\theta}{-9.81}) [/tex]

By equating the time of flight with the distance traveled by the projectile which can be found by dividing the distance moved by the projectile over its horizontal velocity, you can solve for [tex] \theta [/tex]
 

1. How can a garden hose be used to solve physics problems?

A garden hose can be used to demonstrate concepts such as pressure, fluid dynamics, and Bernoulli's principle. By manipulating the flow rate, shape, and direction of the water coming out of the hose, one can observe and analyze the physics behind the movement of the water.

2. What physics principles are involved in using a garden hose to solve problems?

Some of the key principles involved in using a garden hose to solve physics problems include conservation of energy, conservation of mass, and Newton's laws of motion. These principles help explain the behavior of the water as it flows through the hose.

3. How can a garden hose be used to demonstrate Bernoulli's principle?

Bernoulli's principle states that as the speed of a fluid increases, the pressure decreases. This can be demonstrated by turning on a garden hose and pointing it upwards. As the water travels up and out of the hose, the speed increases and the pressure decreases, causing the water to shoot out further.

4. What are some common physics problems that can be solved using a garden hose?

Some common physics problems that can be solved using a garden hose include calculating the velocity or acceleration of the water as it flows through the hose, determining the rate of change of pressure, and analyzing the forces acting on the water.

5. How can using a garden hose to solve physics problems benefit students?

Using a garden hose to solve physics problems can provide students with a hands-on and visual understanding of complex concepts. It can also help them develop critical thinking and problem-solving skills, as they must apply their knowledge of physics principles to real-world situations.

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