A fire hose held near the ground shoots water at a speed of 5.5 .At what angle(s) should the nozzle point in order that the water land 2.0 away (see the figure )?
Mastering physics will take this answer: --> 26,64 <-- degrees.
To solve this problem, we can use the equation for the range (horizontal distance) of a projectile:
R = V^2 * sin(2θ) / g
Where:
R is the range (2.0 m in this case),
V is the initial velocity of the water (5.5 m/s),
θ is the angle at which the nozzle is pointed, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Let's calculate the angle θ:
2.0 = (5.5^2 * sin(2θ)) / 9.8
Multiply both sides by 9.8:
19.6 = 30.25 * sin(2θ)
Divide both sides by 30.25:
sin(2θ) = 19.6 / 30.25
Take the inverse sine of both sides:
2θ = sin^(-1)(19.6 / 30.25)
Now, divide both sides by 2:
θ = (1/2) * sin^(-1)(19.6 / 30.25)
Using a calculator, we find that θ ≈ 22.849 degrees.
Therefore, the nozzle should be pointed at an angle of approximately 22.849 degrees in order for the water to land 2.0 m away.
To determine the angle at which the nozzle should point, we need to consider the motion of the water being shot from the fire hose.
The most important quantity to analyze is the initial velocity of the water. Given that the speed of the water is 5.5 m/s, this represents the magnitude of the initial velocity vector (V0) of the water.
We also know that the water lands 2.0 m away from the nozzle (horizontal displacement). This gives us one component of the initial velocity vector, which is the horizontal component (V0x).
We can use the formula for horizontal displacement to find V0x:
displacement (d) = velocity (v) * time (t)
2.0 m = V0x * t (since the horizontal velocity remains constant)
Next, we need to consider the vertical component of the initial velocity vector (V0y). Since we're dealing with projectile motion, we know that the only force acting in the vertical direction is gravity. Therefore, the initial vertical velocity (V0y) should be determined by the height of the water hose above the ground.
To find the angle at which the nozzle should point, we need to use trigonometry. The tangent function is particularly useful here since it relates the two components of the initial velocity vector.
tan(θ) = V0y / V0x
Now, we can substitute the values we have:
tan(θ) = V0y / V0x = (acceleration due to gravity * t) / V0x
We know that the acceleration due to gravity is approximately 9.8 m/s², and we can substitute V0x from the earlier equation. Solving for θ, we get:
θ = arctan((acceleration due to gravity * t) / 2)
Let's compute this value.
Dx = Vo^2*sin(2A)/g = 2
(5.5)^2*sin(2A)/9.8 = 2,
30.25*sin(2A)/9.8 = 2,
Multiply both sides by 9.8:
30.25*sin(2A) = 19.6,
sin(2A) = 19.6 / 30.25 = 0.64793,
2A = 40.4 Deg.
A = 20.2 Deg.