Water from a garden hose that is pointed 29° above the horizontal lands directly on a sunbather lying on the ground 4.7 m away in the horizontal direction. If the hose is held 1.4 m above the ground, at what speed does the water leave the nozzle?
The time that each drop of water remains in the air is
t = (4.7 m)/V cos 29 .
That is based upon how far it travels
The time t must also satissfy
1.4 + (V sin 29) t - (g/2) t^2 = 0
That is based upon the vertical equation of motion.
Substitute t from the first equation into the second equation and solve for V.
To find the speed at which water leaves the nozzle, we can use the principle of projectile motion.
First, let's break down the given information:
- The water from the hose is pointed 29° above the horizontal.
- The distance between the sunbather and the hose is 4.7 m horizontally.
- The hose is held 1.4 m above the ground.
We can start by determining the initial velocity of the water horizontally and vertically.
Horizontal Component:
The horizontal component of the initial velocity (Vx) remains constant throughout the trajectory because there are no horizontal forces acting on the water once it leaves the nozzle. Since there is no acceleration horizontally, the horizontal component can be calculated using the formula:
Vx = Dx / t,
where Dx is the horizontal distance (4.7 m) and t is the time of flight.
Vertical Component:
The vertical component of the initial velocity (Vy) changes due to the gravity acting vertically downwards. The equation for calculating the vertical component is:
Vy = Vo * sin(θ),
where Vo is the initial velocity and θ is the angle above the horizontal (29°).
Next, we can calculate the time of flight (t) using the vertical component equation and the known values of the vertical displacement (Δy) and the acceleration due to gravity (g = 9.8 m/s²):
Δy = Vo * sin(θ) * t - (1/2) * g * t².
In our case, the vertical displacement is the difference between the initial height of the water from the ground (1.4 m) and the final height of the sunbather (0 m).
Now, we have the values for Vo and t, and we can use these values to calculate Vx using the horizontal component equation.
Finally, we can calculate the speed (v) at which the water leaves the nozzle using the horizontal and vertical components:
v = sqrt(Vx² + Vy²).
By following this step-by-step process, we can calculate the speed at which the water leaves the nozzle.