How Does Projectile Motion Determine Firefighting Hose Angles?

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The discussion focuses on how projectile motion applies to firefighting hose angles when water is shot at a building. Firefighters use a high-pressure hose with an initial water speed of 25.0 m/s, and the water travels 45.0 m in 3.00 seconds. To determine the angle of elevation, the horizontal and vertical components of the water's motion must be analyzed using projectile motion equations. The speed at the highest point of the trajectory can be calculated, where the vertical component becomes zero, while the acceleration remains constant at -9.8 m/s² due to gravity. Understanding these concepts is essential for solving the problems related to the hose's angle and water trajectory.
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Firemen are shooting a stream of water at a burning building using a high-pressure hose that shoots out the water with a speed of 25.0 m/s as it leaves the end of the hose. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation \alphaof the hose until the water takes 3.00 s to reach a building 45.0 m away. You can ignore air resistance; assume that the end of the hose is at ground level

A.) Find the angle of elevation \alpha
B.) Find the speed of the water at the highest point in its trajectory.
C.) Find the acceleration of the water at the highest point in its trajectory(magnitude of the acceleration).

please help me start this, i have problems starting questions.

what is "projectile motion"
25.0 m/s is the initial velocity right? or is it the final velocity?

for problem A.) how would i find the angle?
 
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what is "projectile motion"

Your text should give you a good explanation of this basic concept. You can also try this page I googled just now.

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Vectors/ProjectilesMotion.html

25.0 m/s is the initial velocity right? or is it the final velocity?

the problem stated: "...shoots out the water with a speed of 25.0 m/s as it leaves the end of the hose."

"as it leaves the end of the hose" would imply that it is the initial velocity. But just as an aside, it would also be the final velocity (don't worry about that right now though, you'll learn that when you get to energy conservation.)

for problem A.) how would i find the angle?

Since the water is undergoing projectile motion, you can simplify the problem by considering the horizontal and vertical components of its velocity. I don't want to give the solution away, but think of the data you are given in the problem: a distance, a velocity, and a time. Consider the horizontal component of the velocity using a little trig, and what you now know about projectile motion. This is all i'll tell you for now; draw out a diagram and think about it; if you still have difficulty, post again.
 
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To start, we need to understand what is meant by "projectile motion." Projectile motion refers to the motion of an object that is launched into the air and then moves under the influence of gravity alone. In this case, the water being shot from the hose is the projectile and it will follow a parabolic path under the influence of gravity.

Next, we need to identify what information we have been given and what we need to find. We are given the initial velocity of the water as it leaves the hose (25.0 m/s) and the distance it travels (45.0 m) in a certain amount of time (3.00 s). We are asked to find the angle of elevation, the speed at the highest point, and the acceleration at the highest point.

To find the angle of elevation, we can use the equation for projectile motion: d = v0t + (1/2)at^2, where d is the distance traveled, v0 is the initial velocity, t is the time, and a is the acceleration (in this case, due to gravity). We know the distance (45.0 m) and the time (3.00 s), and we can solve for the initial velocity (v0) using the given speed of 25.0 m/s. This will give us the initial vertical component of the velocity. We can then use trigonometry to find the angle of elevation.

To find the speed at the highest point, we can use the equation v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time. At the highest point, the vertical component of the velocity will be zero, so we can solve for the final velocity (v) using the initial velocity (v0) and the acceleration (a) due to gravity.

To find the acceleration at the highest point, we can use the equation a = -g, where a is the acceleration and g is the acceleration due to gravity (9.8 m/s^2). This will give us the magnitude of the acceleration at the highest point.

I hope this helps you get started with the problem. Remember to always identify what information you have been given, what you are asked to find, and what equations or concepts can help you solve the problem.
 
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