I HONESTLY KNOW THAT THIS SITE IS NOT INTENDED TO JUST GIVE ANSWERS, BUT I HAVE NO IDEA HOW TO DO THIS PROBLEM WHATSOEVER, SO PLEASE MAKE NO COMMNENTS LIKE "post some work" OR ANYTHING LIKE THAT...IF I HAD WORK TO POST....I WOULDN'T REALLY NEED HELP...
:) Thanks.
Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 21.0 m/s as it leaves the hose nozzle. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation of the hose until the water takes 3.00 s to reach a building 35.0 m away. You can ignore air resistance; assume that the end of the hose is at ground level.
Find the angle of elevation of the hose.
Don't know...
Find the speed of the water at the highest point in its trajectory.
Don't know...
Find the acceleration of the water at the highest point in its trajectory.
Don't know...
How high above the ground does the water strike the building?
Don't know...
How fast is it moving just before it hits the building?
Don't know...
You should know the answer to the thrid part at least, since the water follows projectile motion. All projectiles accelerate toward the ground with rate equal to g.
The time that it takes for the water to reach the fire, and the distance D, can be used to tell you the horizontal component of the water velocity leaving the hose, Vo cos A.
(Vo cos A) * (3 seconds) = D
Since the exit velocity Vo is 21 m/s, solve for the cosine of the elevation angle A. From that, get A itself.
At the highest point of the trajectory, the vertical component of velocity is zero, and the horizontal component rmeains the initial value.
Use the initial vertical component Vo sin A to compute how high it goes. It will spend t = 1.50 seconds going up. The height reached is
H = Vo sin A * t - (1/2) gt^2.
For the last part (final velocity) think about applying conservation of energy. The potential energy has not changed.
Please use these hints to make progress on the problem.
To find the angle of elevation of the hose, we can use the given information that the water takes 3.00 seconds to reach a building 35.0 m away.
We can use the horizontal component of the water velocity, Vo cos A, and the time, 3.00 seconds, to calculate the horizontal distance the water travels:
Horizontal distance = (Vo cos A) * (3.00 seconds)
Since the exit velocity Vo is given as 21.0 m/s, we can solve for the cosine of the elevation angle A. From that, we can find A itself.
Next, to find the speed of the water at the highest point in its trajectory, we can look at the horizontal component of velocity. When the water reaches the highest point, its vertical component of velocity becomes zero, while the horizontal component remains constant. So, the speed at the highest point would be the same as the horizontal component of velocity, Vo cos A.
To find the acceleration of the water at the highest point in its trajectory, we know that all projectiles accelerate towards the ground with a rate equal to the acceleration due to gravity, g.
For the height above the ground where the water strikes the building, we can consider the vertical displacement of the water. We can use the initial vertical component of velocity, Vo sin A, and the time spent going up, 1.50 seconds, to calculate the height reached using the equation:
Height = (Vo sin A) * (1.50 seconds) - (1/2) * g * (1.50 seconds)^2
Finally, to find the speed of the water just before it hits the building, we can apply conservation of energy. The potential energy of the water has not changed, so we can equate the initial potential energy to the final kinetic energy:
Initial potential energy = Final kinetic energy
m * g * height = (1/2) * m * (final velocity)^2
Since the mass of the water cancels out, we can solve for the final velocity.