Water spouts from a horizontal pipe 12 meters above the ground. Three meters below the line of the pipe, the water trajectory is at a horizontal distance of 5 meters from the water outlet. How far from the water outlet will the stream of the water hit the ground?
h = 0.5g*t^2 = 3 m.
4.9t^2 = 3
t^2 = 0.612
Tf = 0.782 s. = Fall time.
Dx = Xo*Tf = 5 m.
Xo * 0.782 = 5
Xo = 6.39 m/s = Hor. component of initial velocity.
h = 0.5g*t^2 = 12 m.
4.9t^2 = 12
t^2 = 2.45
Tf = 1.56 s.
Dx = Xo*Tf = 6.39m/s * 1.56s. = 10 m.
From water outlet.
To calculate the distance from the water outlet where the stream will hit the ground, we can use the concept of projectile motion.
Let's consider the vertical and horizontal components of the water's velocity separately.
1. Vertical component: The water is affected by gravity and its vertical motion follows a free-fall trajectory. The initial vertical velocity (Vy) is 0 m/s, and the acceleration due to gravity (g) is approximately 9.8 m/s².
Using the formula:
Δy = Vy * t + (1/2) * g * t²
Where:
Δy = Vertical distance traveled
Vy = Vertical component of velocity
t = Time
We can rearrange the equation to solve for t:
t = √(2 * Δy / g)
The vertical distance traveled, Δy, is 12 meters - 3 meters = 9 meters (since the water outlet is 12 meters above the ground, and the trajectory is 3 meters below the line of the pipe).
Substituting the values into the equation:
t = √(2 * 9 / 9.8)
t ≈ 1.38 seconds
2. Horizontal component: Since the water is traveling horizontally, the horizontal velocity (Vx) remains constant. We are given the horizontal distance (x) as 5 meters.
Using the formula:
x = Vx * t
We can rearrange the equation to solve for Vx:
Vx = x / t
Substituting the values into the equation:
Vx = 5 / 1.38
Vx ≈ 3.62 m/s
Now, we can calculate the horizontal distance traveled by the water to reach the ground.
Using the formula:
Distance = Vx * t
Substituting the values into the equation:
Distance = 3.62 * 1.38
Distance ≈ 4.99 meters
Therefore, the stream of water will hit the ground approximately 4.99 meters away from the water outlet.
To determine how far from the water outlet the stream of water will hit the ground, we can utilize the principles of projectile motion. By breaking down the problem into vertical and horizontal components, we can calculate the distance.
Let's consider the vertical motion first. The problem states that the water spouts from a horizontal pipe 12 meters above the ground. We can calculate the time it takes for the water to hit the ground using the formula for freefall motion:
h = (1/2) * g * t^2
where:
h = height (12 meters)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Plugging the values into the equation, we get:
12 = (1/2) * 9.8 * t^2
Simplifying the equation:
24 = 9.8 * t^2
Now, solve for t:
t^2 = 24 / 9.8
t^2 ≈ 2.449
Taking the square root of both sides:
t ≈ 1.566 seconds
So, it takes approximately 1.566 seconds for the water to hit the ground in the vertical direction.
Next, let's consider the horizontal motion. The problem states that 3 meters below the line of the pipe, the water trajectory is at a horizontal distance of 5 meters from the water outlet.
Since the horizontal velocity remains constant throughout the trajectory, we can calculate it using the formula for horizontal distance:
d = v * t
where:
d = horizontal distance (5 meters)
v = horizontal velocity (which remains constant)
t = time (1.566 seconds)
Rearranging the equation, we can solve for v:
v = d / t
v = 5 / 1.566
v ≈ 3.193 m/s
Now, we can determine how far from the water outlet the stream of water will hit the ground. The horizontal distance traveled by the water can be calculated using the formula:
d = v * t
where:
d = horizontal distance
v = horizontal velocity (3.193 m/s)
t = time (1.566 seconds)
Plugging in the values, we get:
d = 3.193 * 1.566
d ≈ 5.001 (rounded to 3 decimal places)
Therefore, the stream of water will hit the ground approximately 5 meters away from the water outlet.