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 {\rm m} above the ground (see the figure ). When you quickly move the nozzle away from the vertical, you hear the water striking the ground next to you for another 2.2 s.
4 answers
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I assume you want to know how high the stream of water reached.
Well, you have so little information, the useful formulas are necessarily few. How about
s = 1/2 at2 ?
You have a and t, so plug 'em in to get s, the height
s = 4.9 (2.2)2 = 23.716m
That's some nozzle! 7 stories high!
Well, you have so little information, the useful formulas are necessarily few. How about
s = 1/2 at2 ?
You have a and t, so plug 'em in to get s, the height
s = 4.9 (2.2)2 = 23.716m
That's some nozzle! 7 stories high!
Actually, Steve should have used t = 1.1 s in his calculation. That is the amount of time the water spends coming down from maximum height.
s = v(i)t + (1/2)at²
-1.5 = 2.5 v(i) - 4.9 * 2.5²
v(i) = [4.9 * 2.5² - 1.5] / 2.5
v(i) = 11.65 m/s
The answer can't have more than 2 sig figs ... so the water speed as it leaves the nozzle is 12 m/s
-1.5 = 2.5 v(i) - 4.9 * 2.5²
v(i) = [4.9 * 2.5² - 1.5] / 2.5
v(i) = 11.65 m/s
The answer can't have more than 2 sig figs ... so the water speed as it leaves the nozzle is 12 m/s
1 answer