A sprinkler mounted on the ground sends out a jet of water at a 20 ∘ 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?
See previous post: Fri, 10-16-15, 12:41 AM.
Probably
To find out how far the water travels before hitting the ground, we can use the equations of projectile motion.
First, we need to find the time it takes for the water to hit the ground. We can use the formula for time of flight:
T = (2 * u * sinθ) / g
where T is the time of flight, u is the initial velocity of the water, θ is the angle of projection, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values, we get:
T = (2 * 11 * sin(20)) / 9.8
T = (22 * sin(20)) / 9.8
T ≈ 1.33 s (approximately)
Now that we know the time of flight, we can find the horizontal distance traveled by using the formula:
R = u * cosθ * T
where R is the horizontal distance traveled.
Plugging in the values, we get:
R = 11 * cos(20) * 1.33
R ≈ 11 * 0.9397 * 1.33
R ≈ 14.62 m
Therefore, the water travels approximately 14.62 meters before hitting the ground.