A sprinkler mounted on the ground sends out a jet of water at a 30 ∘ angle to the horizontal. The water leaves the nozzle at a speed of 11 m/s . How far does the water travel before it hits the ground?
Dx = Vo^2*sin(2A)/g.
Vo = 11 m/s.
A = 30o
g = 9.8 m/s^2.
Dx = ?
To find the horizontal distance traveled by the water before it hits the ground, we can use the horizontal component of the velocity.
Given that the water leaves the nozzle at a speed of 11 m/s and at an angle of 30 degrees to the horizontal, we can find the horizontal component of the velocity using trigonometry.
The horizontal component of the velocity (v_x) is given by:
v_x = v * cosθ,
Where v is the initial velocity (11 m/s) and θ is the angle of the jet of water (30 degrees).
v_x = 11 m/s * cos(30 degrees)
Using the cosine of 30 degrees (cos(30) = √3/2), we can substitute it back into the equation:
v_x = 11 m/s * √3/2
v_x = 11/2 * √3
Now, we need to find the time (t) it takes for the water to hit the ground. We can use the formula:
t = (2 * v * sinθ) / g,
Where g is the acceleration due to gravity (9.8 m/s^2).
t = (2 * 11 m/s * sin(30 degrees)) / 9.8 m/s^2
Using the sine of 30 degrees (sin(30) = 1/2), we can substitute it back into the equation:
t = (2 * 11 m/s * 1/2) / 9.8 m/s^2
t = (11 m/s) / (9.8 m/s^2)
Simplifying further:
t ≈ 1.122 seconds
Now, we can calculate the horizontal distance (d) using the equation:
d = v_x * t
Plugging in the values:
d = (11/2 * √3) m/s * 1.122 s
d ≈ 6.04 meters
Therefore, the water travels approximately 6.04 meters before it hits the ground.
To find the horizontal distance traveled by the water before it hits the ground, we need to analyze the vertical and horizontal components of the water's velocity separately.
Given:
- Angle of the water jet to the horizontal, θ = 30°
- Speed of the water, v = 11 m/s
First, we need to find the vertical component of the water's velocity. This can be calculated using the formula:
v_y = v * sin(θ)
where v_y is the vertical component of the velocity.
Substituting the given values:
v_y = 11 m/s * sin(30°)
v_y = 11 m/s * 0.5
v_y = 5.5 m/s
Next, we can find the time it takes for the water to hit the ground by considering the vertical motion. The equation for the vertical displacement is given by:
y = v_y * t + (1/2) * g * t^2
where y is the vertical displacement, t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s^2). Since the water hits the ground, we can set y = 0.
0 = 5.5 m/s * t + (1/2) * 9.8 m/s^2 * t^2
Rearranging the equation:
4.9 t^2 + 5.5 t = 0
Using the quadratic formula to solve for t, we get:
t = (-5.5 ± √(5.5^2 - 4 * 4.9 * 0)) / (2 * 4.9)
t = (-5.5 ± √(30.25)) / 9.8
Since time cannot be negative, we take the positive root:
t ≈ 0.76 seconds
Now that we know the time it takes for the water to hit the ground, we can find the horizontal distance traveled using the equation:
x = v_x * t
where v_x is the horizontal component of the velocity.
The horizontal component can be calculated using the formula:
v_x = v * cos(θ)
Substituting the given values:
v_x = 11 m/s * cos(30°)
v_x = 11 m/s * √(3)/2
v_x = 9.53 m/s
Finally, substituting the values of v_x and t into the equation for horizontal distance:
x = 9.53 m/s * 0.76 s
x ≈ 7.25 meters
Therefore, the water travels approximately 7.25 meters horizontally before hitting the ground.