A gardener holds a hose 0,75m above the ground such that the waterfront out horizontally and hits the ground at a point 2 meters away, what is the speed with which the water leaves the hose.
123.56
How long does it take the water to hit the ground?
4.9t^2 = 0.75
t = 0.3912 seconds
so, the horizontal speed =
2m/.3912s = 5.11 m/s
G: dy=0.75m dx=2.0m
F: Vix=?
S: t=√2dy/g
=√2•0.75/9.8
=√1.5/9.8
=1.225/3.13
=0.391s
Vix=dx/t from dx=Vixt
=2.0÷0.391
=5.115m/s
To determine the speed with which the water leaves the hose, we can use the principles of projectile motion. We can use the horizontal distance (2 meters) and the height of the hose from the ground (0.75 meters) to calculate the initial velocity of the water.
First, let's assume that there is no air resistance affecting the water. In this case, we can assume that when the water leaves the hose, it follows a projectile motion trajectory, similar to a thrown object.
The key concept to understand is that the horizontal and vertical motions of the water are independent of each other. This means we can treat them separately.
First, let's calculate the time it takes for the water to reach the ground horizontally. We know the horizontal distance (2 meters) and we can assume there is no horizontal acceleration, so we can use the following formula:
Distance = Velocity * Time
Since the velocity in the horizontal direction remains constant, we can rearrange the formula to solve for time:
Time = Distance / Velocity
In this case, the distance is 2 meters and the velocity in the horizontal direction is the same as the initial velocity of the water. Therefore, the time it takes for the water to reach the ground horizontally is:
Time = 2 meters / Velocity
Now, let's calculate the vertical distance the water travels using the equation for vertical motion:
Vertical distance = (Initial vertical velocity * Time) - (0.5 * Acceleration due to gravity * Time^2)
Here, the initial vertical velocity is 0 (since the water is not moving vertically when it leaves the hose), the time is the same as the previous calculation, and the acceleration due to gravity is approximately 9.8 m/s^2 (assuming Earth's gravity).
The vertical distance is 0.75 meters, so we can plug in the values and solve for the time:
0.75 meters = (0 * Time) - (0.5 * 9.8 m/s^2 * Time^2)
Simplifying the equation, we get:
4.9 * Time^2 = 0.75 meters
Solving for Time, we find:
Time = √(0.75 meters / 4.9 m/s^2)
Now that we have the time, we can substitute it back into our earlier equation to solve for the velocity:
Velocity = Distance / Time
Using the given distance of 2 meters and the calculated time, we can find the velocity:
Velocity = 2 meters / Time
By substituting the calculated value of Time into this equation, we can find the velocity with which the water leaves the hose.