The nozzle of a fountain jet sits in the center of a circular pool of radius 3.50 m. If the nozzle shoots water at an angle of 65 degrees.
A) What is the maximum speed of the water at the nozzle that will allow it to land within the pool?
To find the maximum speed of the water at the nozzle that will allow it to land within the pool, we can consider the projectile motion of the water.
Given:
Radius of the circular pool, r = 3.50 m
Angle at which the water is shot, θ = 65 degrees
To solve this problem, we need to consider the horizontal and vertical components of the water's motion separately.
First, let's find the horizontal component of the water's velocity at the nozzle. Since there is no acceleration horizontally, the horizontal component of the velocity remains constant throughout the motion. We can use the cosine function to find the horizontal component of the initial velocity (Vx):
Vx = V * cos(θ)
Next, let's find the vertical component of the water's velocity at the nozzle. The water is accelerated downward due to gravity, so we need to consider the vertical component of the velocity (Vy) by using the sine function:
Vy = V * sin(θ)
In order for the water to land within the pool, the horizontal distance it travels should be equal to or less than the diameter of the pool. Since the diameter is twice the radius (2r), we can write the following equation:
2r ≥ Vx * t
Here, t represents the time it takes for the water to reach the ground.
Now, we need to find the value of V (water's initial velocity) that satisfies the condition. Since V is the initial velocity at the nozzle and we want to find the maximum speed, we need to consider the highest possible value of V.
Let's consider the maximum vertical distance the water can travel within the pool. This occurs when the water is shot at an angle of 45 degrees. At this angle, the vertical component of the velocity (Vy) is the same as the value of V.
Using Vy = V * sin(θ) and considering θ as 45 degrees, we have:
Vy = V * sin(45)
Since sin(45) = 1/sqrt(2), we can rewrite the equation as:
Vy = V/sqrt(2)
We also know that the time it takes for an object to reach maximum height and then come back down is given by:
t = 2 * (Vy/g)
Here, g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting Vy = V/sqrt(2) into the equation, we get:
t = 2 * (V/sqrt(2)) / g
Now, let's substitute this value of t into our initial equation:
2r ≥ Vx * [2 * (V/sqrt(2)) / g]
Simplifying this equation, we find:
2r ≥ (V * cos(θ)) * [2 * (V/sqrt(2)) / g]
2r ≥ (V^2 * 2 * cos(θ)) / (sqrt(2) * g)
2r ≥ (2 * V^2 * cos(θ)) / (sqrt(2) * g)
We want to find the maximum value of V, so let's consider the equality condition:
2r = (2 * Vmax^2 * cos(θ)) / (sqrt(2) * g)
Simplifying further:
Vmax^2 = (2r * sqrt(2) * g) / (2 * cos(θ))
Finally, to find the maximum speed (Vmax):
Vmax = sqrt((2r * sqrt(2) * g) / (2 * cos(θ)))
Plugging in the given values:
Vmax = sqrt((2 * 3.50 * sqrt(2) * 9.8) / (2 * cos(65)))
Calculating this expression will give you the maximum speed of the water at the nozzle that will allow it to land within the pool.