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

In summary: It seems you have figured out the mistake. In summary, the problem involves firemen using a high-pressure hose to shoot water at a burning building. The water moves in projectile motion and the firemen adjust the angle of elevation until it takes 3 seconds to reach a building 45 meters away. The angle of elevation is 53.1 degrees and the speed of the water at the highest point is 15 m/s with an acceleration of 9.8 m/s^2. The water strikes the building at a height of 15.9 meters with a velocity of 17.7 m/s. However, previous attempts at using equations such as y = (voy+vy)/2 * t and vy^2 = voy^2
  • #1
Toranc3
189
0

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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  • #2
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.
 
  • #3
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.
 
  • #4
No need to try it anymore. I've got it now. Thank you nasu.
 
  • #5
You are welcome.
 

1. What is projectile motion?

Projectile motion is the motion of an object through the air, under the influence of gravity. It is a combination of horizontal and vertical motion, resulting in a parabolic path.

2. How is projectile motion calculated?

The motion of a projectile can be calculated using the equations of motion, which take into account the initial velocity, acceleration due to gravity, and time. These equations can be used to determine the position, velocity, and acceleration of the object at any given time.

3. Why doesn't checking an answer with another equation always work for projectile motion?

While projectile motion can be calculated using multiple equations, it is important to remember that these equations only apply in ideal conditions, such as a vacuum. In the real world, there are external factors such as air resistance that can affect the motion of the object, making the equations less accurate.

4. What are some common mistakes when solving projectile motion problems?

One common mistake is not properly considering the direction of the initial velocity. Another mistake is using the wrong equation for the given scenario, such as using an equation for horizontal motion when the problem involves both horizontal and vertical motion.

5. How is projectile motion used in real life?

Projectile motion has many practical applications, such as in sports like basketball and baseball, where players need to accurately throw or shoot objects. It is also used in engineering and physics to study the motion of objects and design structures such as bridges and buildings.

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