3 Physics Problem: Please Help

  • Thread starter mt2568
  • Start date
  • #1
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?
 

Answers and Replies

  • #2
Problems 1 and 2 have been answered, but number three still alludes me...
 
  • #3
eludes not alludes, hehe
 
  • #4
enigma
Staff Emeritus
Science Advisor
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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.
 
  • #5
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
 
  • #6
Actually forgot to divide that 2s, its 9.6 m/s
 
  • #7
Also forgot that one part is .5 not 2, i wrote it wrong but still calculated with the .5
 
  • #8
Antiproton
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0
I also got 9.65 (remember those sig figs!!!). Congratulations on this problem. Keep up the hard work.
 
  • #9
Isn't the second kinematic equation (1/2)AT^2 instead of 2AT^2?
 
  • #10
Antiproton
39
0
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.
 

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