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SSM A fire hose ejects a stream of water at an angle of 35.0 above the horizontal. The water leaves the nozzle with a speed of 25.0 m/s. Assuming that the water behaves like a projectile, how far from a building should the fire hose be located to hit the highest possible fire?

$30,0 m$

Physics 101 Mechanics

Chapter 3

Kinematics in Two Dimensions

Motion in 2d or 3d

Cornell University

Hope College

University of Sheffield

University of Winnipeg

Lectures

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So the question states that a water jet is shot out 25 meters per second at a height at an angle of 35 degrees. And we're trying to find how far away this water jet needs to be so that it reaches a maximum height on this building a distance X away. So to do this, we really need a first find out how long it takes for the water jet to reach its maximum height. Um and so we can break up this Fosse into a vertical and horizontal component so we can figure out the vertical component because we know that sign of 35 degrees is equal to opposite over high pot news, which means that the vertical component is equal to 25 times sign of 35 degrees. And now that we know the vertical component of velocity, we can use our kingdom attics equation. So in this case, we're gonna use the equation that states that the final boss he's equal to B and Michelle velocity was the acceleration times time we knew that the fun of lost E at its maximum height is going to be zero in the vertical vertical direction. So be zero is equal to the initial vertical glossy 25 times sign of 35 degrees, plus the acceleration, which is negative 9.8 meters per second squared because of gravity times time. So once we do this, we can solve for the time and we'll find out that T is equal to 1.463 seconds. And now that we know this, we can figure out what the horizontal velocity is. The initial horizontal velocity V of X and V of X is just going to be We know that co sign of 35 degrees is the Jason Overhype on U so v of X, over 25 meters per second, which means V of X is equal to 25 co sign of 35 degrees. And now, since we know the horizontal velocity and we also time, we know that distance, uh, the jet will travel to reach its maximum height in the distance in the X direction and so weakened do 25 co sign of 35 degrees times the time, which is one point for 63 seconds, and this will give us our Delta X. And when we do this calculation, we find that Delta X is equal to 30 meters

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