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?
Write vertical and horizontal equations of motion, with the time of flight t and the initial velocity V as the two unknowns. Solve the two equations in two unknowns.
V*cos 29*t = 0.8746 V * t = 4.7
V*sin 29*t - (g/2) t^2 =
= 0.4848 V - 0.49 t^2 = -1.4
Substitute 0.5374/V for t in the second equation, and you will end up with a single equation for V that you must solve.
To find the speed at which the water leaves the nozzle, we can use the principles of projectile motion. We need to break the velocity of the water into its horizontal and vertical components.
Let's start by analyzing the horizontal motion of the water. Since there are no horizontal forces acting on the water, the horizontal velocity remains constant throughout its path. This means the horizontal component of the water's velocity is equal to the horizontal component of the initial velocity.
The horizontal component of the velocity (Vx) can be found using the equation:
Vx = Velocity * cosθ
where θ is the angle at which the water is pointed above the horizontal.
In this case, θ = 29°. Let's calculate the horizontal component of the velocity:
Vx = Velocity * cos(29°)
Now, let's analyze the vertical motion of the water. The water is accelerated downward due to gravity. The vertical component of the velocity decreases with time until the water hits the ground. The time it takes for the water to hit the ground can be found using the equation:
t = sqrt((2 * h) / g)
where h is the initial vertical displacement (1.4 m) and g is the acceleration due to gravity (9.8 m/s²).
With the time (t) known, we can find the vertical component of the velocity (Vy) using the equation:
Vy = g * t
Finally, we can find the initial velocity (V) using the equation:
V = sqrt(Vx² + Vy²)
Now, let's put it all together and calculate the initial velocity:
1. Calculate the horizontal component of the velocity:
Vx = Velocity * cos(29°)
2. Calculate the time taken to hit the ground:
t = sqrt((2 * 1.4) / 9.8)
3. Calculate the vertical component of the velocity:
Vy = 9.8 * t
4. Calculate the initial velocity:
V = sqrt(Vx² + Vy²)
By plugging in the values into these equations, we can determine the speed at which the water leaves the nozzle.