Projectile moting and checking an answer with another equation doesn't work.

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The discussion centers on a physics problem involving projectile motion, where water is shot from a hose at a building. The calculations for the angle of elevation, speed, and height of the water are presented, with the angle found to be 53.1 degrees and the height at impact calculated as approximately 15.9 meters. The user expresses confusion about discrepancies when using different equations to find the height, particularly when assuming the water hits the building at the peak of its trajectory. Responses highlight the importance of clarity in problem assumptions and the necessity of including units in calculations. Ultimately, the user resolves their confusion with assistance from another participant.
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Homework Statement



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 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 of the hose until the water takes 3.00s to reach a building 45.0 away. You can ignore air resistance; assume that the end of the hose is at ground level.

a. Find the angle of elevation
b. Find the speed and acceleration of the water at the highest point in its trajectory
c.how high above the graound does the water strike the building and how fast is it moving just before it hits the building?

Homework Equations



x-xo=vox*t
vx=vox
y=yo +voy*t -1/2 *g *t^(2)
vy=voy-g*t
y-yo=(voy+vy)/2 *t
vy^(2)=voy^(2) -2g(y-yo)

The Attempt at a Solution



a= 53.1 degrees
x=xo+vox*t
45.0m=25.0m/s*cos(theta)*3.00s
Theta=53.1 degrees

b=15 m/s and g=9.8m/s^(2)
Vox=25.0m/s*cos(53.1)= 15m/s
Aceleration is the same vertically.

c=15.9m
y=yo+voy*t -1/2*g*t^(2)
y= 20m.s*3.00s-1/2*g*9.00s^(2)
y=15.8m

d=17.7m/s
vy^(2)=voy^(2)-2g(y-yo)
vy^(2)=(20m.s)^(2)-19.6m/s^(2)*15.9m
vy=9.4m/s

therefore Vf=17.7m/s by v=sqrt(vx^(2)+(vy)^(2))
vx=15.0m/sMy question is is that when I try to check my answer with other equations they do not come out the same. Why is that?
For example: I want to get my heigth at the highest point

vy^(2)=voy^(2)-2g(y-yo)
0=(20m/s))^(2)-19.6*y i get y=20m

y-yo=(voy+vy)/2 * t
y-0=(20m/s+0)/2*3.00s here i get 30m

How come these equations won't work to give me the actual height of 15.9m?

Homework Statement


Homework Equations


The Attempt at a Solution


Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
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1. You did not show your solution but just final results. How can someone guess what is wrong?
2. The values without units are useless. It takes 3s or 3 hours to reach the building? The answer will be different for the two cases.
3. Are you assuming that the water hits the building at the maximum point of the trajectory? The problem does not say so.
 
nasu said:
1. You did not show your solution but just final results. How can someone guess what is wrong?
2. The values without units are useless. It takes 3s or 3 hours to reach the building? The answer will be different for the two cases.
3. Are you assuming that the water hits the building at the maximum point of the trajectory? The problem does not say so.

There I fixed it. Yes I did assume that the water hits the building at the max point.
 
No need to try it anymore. I've got it now. Thank you nasu.
 
You are welcome.
 
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