Question

You’re holding a hose at waist height and spraying water horizontally with it. the hose nozzle has a diameter of 1.80 cm, and the water splashes on the ground a distance of 0.950 m horizontally from the nozzle. suppose you now constrict the nozzle to a diameter of 0.750 cm; how far horizontally from the nozzle will the water travel before hitting the ground? (ignore air resistance.)

Answer 1

When the diameter of the nozzle is changed from 1.80 cm to 0.750 cm, the water travels a distance of 0.395 m horizontally before hitting the ground.

Step-by-step explanation:

To calculate the distance the water will travel before hitting the ground, we can use the principle of conservation of energy. The initial potential energy of the water is converted into kinetic energy as it travels horizontally. We can use the equation:

mgh = (1/2)mv^2

Where m is the mass of the water, g is the acceleration due to gravity, h is the initial height, and v is the final velocity of the water. Since the height is at waist level, which is approximately 1 meter, and the final velocity is zero (since the water hits the ground), we can solve for the distance horizontally from the nozzle. Let's solve for the original diameter of 1.80 cm first:

1.80 cm = 0.018 m

Using the equation:

d = v*t

We can calculate the time it takes for the water to travel to the ground:

t = d/v

Substituting the given values:

t = 0.950 m / v

Now, let's substitute the values to solve for the velocity:

mgh = (1/2)mv^2

mg = (1/2)v^2

9.8 m/s^2 = (1/2)v^2

v^2 = 19.6 m^2/s^2

v = sqrt(19.6) m/s

Now substitute the values of v and solve for t:

t = 0.950 m / sqrt(19.6) m/s

Now, we can use the time calculated to solve for the new distance when the diameter is 0.750 cm:

t_new = d_new / v_new

Since the water travels horizontally and the final velocity is the same as before, we can rearrange the equation and solve for d_new:

d_new = t_new * v

Substituting the values, we get:

d_new = 0.950 m * (0.750 cm / 1.800 cm) = 0.395 m