Water from a garden hose that is pointed 23° above the horizontal lands directly on a sunbather lying on the ground 3.9 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?
To find the speed at which the water leaves the nozzle, we can use the principles of projectile motion. Here's how you can approach and solve the problem step by step:
Step 1: Draw a diagram
Start by drawing a diagram to visualize the given information. Draw a horizontal line to represent the ground and mark the position of the sunbather. Draw a line representing the path of the water, which is at an angle of 23° above the horizontal. Label the horizontal distance between the hose and the sunbather as "3.9 m" and the vertical distance between the hose and the ground as "1.4 m."
Step 2: Break down the motion into horizontal and vertical components
The motion of the water can be broken down into two components: horizontal (x-direction) and vertical (y-direction).
The horizontal component of the water's velocity remains constant and is equal to the velocity at which it leaves the nozzle. We can denote this velocity as "Vx."
The vertical component of the water's velocity changes due to gravity. The initial vertical velocity is zero, and the acceleration due to gravity is denoted as "g" (approximately 9.8 m/s²).
Step 3: Solve for time of flight
To find the time it takes for the water to land on the sunbather, we need to calculate the time of flight (t) by using the horizontal distance (3.9 m) and horizontal velocity (Vx). We can use the equation:
Horizontal distance = Horizontal velocity × Time of flight
3.9 m = Vx × t
Solving for t, we get:
t = 3.9 m / Vx
Step 4: Solve for vertical velocity
The vertical distance traveled by the water is the difference between the initial height of the nozzle (1.4 m) and the final height of the sunbather (0 m). We can use the equation of motion:
Vertical distance = (Initial vertical velocity × Time of flight) + (0.5 × acceleration due to gravity × (Time of flight)²
0 m = (0 m/s × t) + (0.5 × 9.8 m/s² × t²)
0 m = 0 m/s × t + 4.9 m/s² × t²
0 = 4.9 m/s² × t²
Step 5: Solve for vertical velocity
Since the equation is quadratic, we know that t must be equal to zero or there are two values of t that satisfy the equation. The time it takes for the water to reach the ground is the positive solution.
Using t = 3.9 m / Vx from Step 3, we can substitute it into the equation:
0 = 4.9 m/s² × (3.9 m / Vx)²
Simplifying the equation, we get:
19.11 m²/s² = (3.9 m / Vx)²
Step 6: Solve for Vx
To find the horizontal velocity (Vx), we can take the square root of both sides of the equation:
√(19.11) m/s = 3.9 m / Vx
Solving for Vx, we get:
Vx = 3.9 m / √(19.11) m/s
Using a calculator, we can find the approximate value of Vx:
Vx ≈ 0.848 m/s
Therefore, the water leaves the nozzle with a speed of approximately 0.848 m/s.