What are these 3 physics problems and can you solve them?

  • Thread starter Thread starter mt2568
  • Start date Start date
  • Tags Tags
    Physics
AI Thread Summary
The discussion revolves around solving three physics problems related to motion and kinematics. The first two problems have been addressed, focusing on the distance traveled by a falling object and the height from which a stone falls past a window. The third problem involves calculating the speed of water exiting a garden hose nozzle, with participants discussing the application of kinematic equations to derive the solution. A correction was made regarding the use of the second kinematic equation, emphasizing the importance of accuracy in calculations. The conversation highlights collaborative problem-solving and encourages sharing work for better assistance.
mt2568
If you know how to do these, please either answer them here or catch me on AIM: Tregaron2002

1. For an object falling freely from rest, show that the distance traveled during each successive second increases in the ration of successive odd integers.

2. A fall stone takes .30 seconds to travel past a window 2.2 m tall. From what height above the top of the window did the stone fall?

3. Suppose you adjust your garden hose nozzle for a hard stream of water. You point the nozzle vertically upward at a height of 1.5 m above the ground (the hose is 1.5 m off the ground). When you quickly move the nozzle from the vertical you hear the water striking the ground next to you for another 2.0 s. What is the water speed as it leaves the nozzle?
 
Physics news on Phys.org
Problems 1 and 2 have been answered, but number three still alludes me...
 
eludes not alludes, hehe
 
Welcome to the forums,

Think of the water's sound stopping as the time when the last of the water finished it's motion.

You can model the last of the water as a particle, then.

You have the total time, angle, starting height, ending height, and gravity. Solve for velocity.

Does that make sense, or do you need more?

BTW: for help in homework on these forums, we ask that you show your work and where you got stuck.
 
This how I got my answer, not sure if it was right though...

v = V0T + 2AT^2 (2nd Kinematic equation)

x-1.5 = v0(2s) + 2(9.8m/s^2)(2s)

When T = 2, then x- 1.5 is zero (When the water hits the gound the velocity of the water is zero)

v0 = -19.6m/s or 19.6 m/s
 
Actually forgot to divide that 2s, its 9.6 m/s
 
Also forgot that one part is .5 not 2, i wrote it wrong but still calculated with the .5
 
I also got 9.65 (remember those sig figs!). Congratulations on this problem. Keep up the hard work.
 
Isn't the second kinematic equation (1/2)AT^2 instead of 2AT^2?
 
  • #10
Yes, it is. But he found his mistake and corrected it. Using 1/2 instead of 2, he got the appropriate answer.
 
  • #11
Oh, right. Heh, sorry, I didn't notice that.
 

Similar threads

Replies
3
Views
2K
Replies
38
Views
3K
Replies
2
Views
2K
Replies
5
Views
2K
Replies
2
Views
8K
Back
Top