A gardener holds a hose 0.75m above the ground such that the water shoots out horizontally and hits the ground at appoint 2.0 m away. What is the speed with which the water leaves the hose?

Question

A gardener holds a hose 0.75m above the ground such that the water shoots out horizontally and hits the ground at appoint 2.0 m away. What is the speed with which the water leaves the hose?

Answer 1

so, how long does it take to fall 0.75 meters?

4.9t^2 = 0.75

Using that time value, the constant horizontal speed is

2.0m/ts = 2.0/t m/s

Answer 2

To find the speed with which the water leaves the hose, we need to understand the relationship between the horizontal distance the water travels and the time it takes to reach that distance.

Step 1: Determine the time it takes for the water to hit the ground.
We can use the fact that the water shoots out horizontally to calculate the time it takes to reach the ground. The vertical distance traveled by the water can be calculated using the equation:

𝑑 = 𝑣0𝑡 + 0.5𝑔𝑡^2

Where:
𝑑 = vertical distance traveled by the water (0.75m)
𝑣0 = initial vertical velocity of the water (0 m/s)
𝑔 = acceleration due to gravity (-9.8 m/s^2)
𝑡 = time taken

Since the initial vertical velocity of the water is 0 m/s (water shoots out horizontally), the equation simplifies to:

0.75 = -4.9𝑡^2

Simplifying further, we have:

0.75/(-4.9) = 𝑡^2

Step 2: Calculate the time it takes for the water to hit the ground.
Solving for 𝑡, we get:

𝑡 = √(0.75/4.9)

𝑡 ≈ 0.319 seconds

Step 3: Calculate the horizontal velocity of the water.
The horizontal distance traveled by the water (2.0 m) is equal to the horizontal velocity (𝑣) multiplied by the time (0.319 s).

𝑣 = 𝑑 / 𝑡

𝑣 = 2.0 / 0.319

𝑣 ≈ 6.27 m/s

Therefore, the speed with which the water leaves the hose is approximately 6.27 m/s.

Answer 3

To find the speed with which the water leaves the hose, we can use the principle of projectile motion. When the water shoots out horizontally, it acts as a projectile in the vertical direction. The time it takes for the water to hit the ground can be found using the vertical motion equation:

y = y0 + v0y * t - (1/2) * g * t^2

where:
y = final vertical displacement (0 m in this case since it hits the ground)
y0 = initial vertical displacement (0.75 m in this case)
v0y = initial vertical velocity (unknown)
g = acceleration due to gravity (-9.8 m/s^2)
t = time (unknown)

Since it is moving horizontally, the horizontal velocity (v0x) is constant throughout the motion.

The horizontal displacement (x) can be found using the horizontal motion equation:

x = v0x * t

where:
x = horizontal displacement (2.0 m in this case)
v0x = initial horizontal velocity (unknown)
t = time (unknown)

Since the water is shot out horizontally, the initial vertical velocity (v0y) is zero.

Using the time (t) obtained from the horizontal and vertical equations, we can find v0x using the relation:

v0x = x / t

Now, we can calculate the v0x:

v0x = 2.0 m / t

Substituting the value of v0x into the vertical equation, we can solve for t:

0 = 0.75 m + 0 * t - (1/2) * (-9.8 m/s^2) * t^2

Simplifying the equation, we get:

(1/2) * 9.8 m/s^2 * t^2 = 0.75 m

Solving for t, we get:

t^2 = (2 * 0.75 m) / 9.8 m/s^2

t^2 = 0.153 m / m/s^2

t = √(0.153 m / m/s^2)

t ≈ 0.39 s

Now, substituting the value of t into the equation for v0x, we get:

v0x = 2.0 m / 0.39 s

v0x ≈ 5.13 m/s

Therefore, the speed with which the water leaves the hose is approximately 5.13 m/s.