Semi-hard Projectile Motion Problem.

In summary, the problem involves a fireman directing a stream of water from a ground level fire hose at an angle of 30 degrees above the horizontal towards a building 50m away. The speed of the stream is 40 m/s. The solution so far involves finding the x and y independent velocity components, the time to the highest point, the total distance traveled, and the highest point reached. However, to find the height at which the stream of water will strike the building, the time it takes for the horizontal displacement of 50m must be found and then used in the equation for the vertical displacement. The correct answer for the height of the building is 18.66 meters.
  • #1
AznBoi
471
0
Ok, I've just come across a confusing projectile motion problem. I have my own solution for it from the work that I have done so far. Please help me finish the problem for the information I have so far. Thanks a lot!:tongue:

Problem: A fireman, 50m away from the burning building, directs a stream of water from a ground level fire hose at an angle of 30 degrees above the horizontal. If the speed of the stream as it leaves the hose is 40 m/s, at what height will the stream of water strike the building?

My solution so far from the work I have done.
Ok, first I found the x and y independent velocity components of 40m/s at 30 degrees above horizontal. Vy=20m/s; Vx=34.64 m/s

I found the time to the highest point, 2.041s; therefore the time of the total flight is 4.082s.

Then I found x (total distance traveled) which is 141.4m

Then I found the highest point which is 20.41m

But the I realized that I had to find the height of when it reaches the building 50 meters away... I have all the information for the TOTAL FLIGHT. How am I suppose to find the height of when the distance is 50m?? Do I substitute 50m for X=Xo+Voxt+.5AxT^2?? Then I find the time of that distance, then the hight?? What am I doing wrong here??

At first, I wanted to find all the info of the TOTAL FLIGHT then try to come up with a coordinate for 50m, but now I realized that it would be hard to graph it. I don't know what to do, I haven't really done a lot of these problems before.


However, I understand the basic concepts of how to find the information of these kind of projectile motion problems. Thanks a lot for you help!:tongue:
 
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  • #2
Since, [tex]x(t)=v_{0x}t[/tex], you can easily find the time it takes to reach the building by plugging in the distance from the building and the x-component of the initial velocity. Further on, use that time in the equation of displacement for the y-direction to obtain the height.
 
  • #3
So is this correct?:

50m=Vxo*t
50m=34.64m/s*t

t=50m/(34.64m/s)

Then you plug the time in for y=Vyot+.5gt^2:

y=20m/s*t+.5(-9.8m/s^2)(t)^2

So that would be the height of where the water hits the building at 50m away?
 
  • #4
Yes, it should be, after you plug in t = 50/34.64 sec.
 
  • #5
Alright, that's what I was thinking of before. I get it now, thanks for your help! :tongue: I wasted all my time finding the total information =P
 
  • #6
you need to find the time when the horizontal displacement is 50, and then find the vertical displacement at that point.

I got t= 5/3
 
  • #7
maaan, you guys reply so fast!
 
  • #8
Ahmedstein said:
you need to find the time when the horizontal displacement is 50, and then find the vertical displacement at that point.

I got t= 5/3
The time would be:

Avg. V= X/t

t=X/V

So the displacement would be 50m and the Avg. velocity or Vxo would be 34.64m/s. I already found the Vxo component in the beginning.

Therefore the time would be: 50m/34.64m/s :tongue:

Btw, the answer that I got for the height of the building that the water hits is: 18.66 meters.

18.66m makes sense because the highest point(total) would be 20.41m. The highest point's time is 2.041 seconds. For the time of 50m I got 1.4434s which would make the 18.66m be pretty accurate.
 
  • #9
Can anyone check to see if 18.66 m is the correct height? Thanks :smile:
 
  • #10
AznBoi said:
The time would be:

Avg. V= X/t

t=X/V

So the displacement would be 50m and the Avg. velocity or Vxo would be 34.64m/s. I already found the Vxo component in the beginning.

Therefore the time would be: 50m/34.64m/s :tongue:
yeah dude, my bad.. but it's not the average velocity it's just the x component of V which is constant at each point. anyway

Y = 20*1.44 - 0.5*9.8*1.44^2
= 18.64 Hooooray
 

1. What is semi-hard projectile motion?

Semi-hard projectile motion refers to a type of motion where an object is launched at an angle with some initial velocity, and then travels through the air under the influence of gravity. The term "semi-hard" is used to indicate that the object is not launched at a perfectly horizontal or vertical angle, but instead at an angle in between.

2. How do you solve a semi-hard projectile motion problem?

To solve a semi-hard projectile motion problem, you will need to use the equations of motion, including those for displacement, velocity, and acceleration. You will also need to consider the initial velocity, angle of launch, and the effects of gravity. It may be helpful to draw a diagram and break the motion into horizontal and vertical components.

3. What are the key assumptions in semi-hard projectile motion?

The key assumptions in semi-hard projectile motion are that there is no air resistance, the acceleration due to gravity is constant, and the object is launched at a constant initial velocity. These assumptions allow for the use of simplified equations to solve the problem.

4. How does the angle of launch affect the motion of a semi-hard projectile?

The angle of launch has a significant impact on the motion of a semi-hard projectile. A lower launch angle will result in a shorter horizontal distance traveled, while a higher launch angle will result in a longer horizontal distance. The angle also affects the maximum height reached by the object and the time of flight.

5. Can semi-hard projectile motion be applied to real-world situations?

Yes, semi-hard projectile motion can be applied to real-world situations, such as the motion of a baseball being thrown, a basketball being shot, or a rocket being launched. However, in most real-world situations, there are additional factors to consider, such as air resistance, which may require more complex equations to accurately model the motion.

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