A sprinkler mounted on the ground sends out a jet of water at a 30-degree angle to the horizontal.
A sprinkler mounted on the ground sends out a jet of water at a 30-degree angle to the horizontal. The water leaves the nozzle at a speed of 12 m/s. How far does the water travel before it hits the ground?
1 Answers
First, we need to break down the component vectors of the velocity vector.
Vy = 13sin30° = 6.5 m/s;
Vx = 13cos30° = 11.2 m/s;
Now, we need to find the time that the water hits the ground by using the y component.
0 = di + vt - (1/2)at² = 6.5t - 4.8t² = t(6.5 - 4.8t); 6.5 - 4.8t = 0, t = 6.5/4.8 = 1.35 sec;
The water has a height of 0 meters at t = 0 and t = 1.35 seconds. Since t = 0 is the initial time, t = 1.35 is the time the water hits the ground.
We can use that time and the x component to find its distance when it hits the ground.
df = di + vt + (1/2)at² = (11.2)(1.35) = 15.1;
The water reached a distance of 15.1 meters before it hit the ground.
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