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Old 10-15-2006, 08:05 PM   #1
hi its me alec
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Bored? Help me with physics!!

1.) A diver running 1.0 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.8 s later. How high was the cliff and how far from its base did the diver hit the water?

a.) _______m (cliff)
b.) _______m (distance from base)

2.) A fire hose held near the ground shoots water at a speed of 7.9 m/s.


(a) At what 2 angles could the nozzle point in order that the water would land 2.0 m away?
_______ (smaller angle)
_______ (larger angle)

(b) Why are there two different angles?

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Old 10-15-2006, 08:18 PM   #2
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hmm for 1 a) you can use x=1/2 at^2...soo its x=-4.9(2.8)^2.
for 1b) you can use x=vavgt so thats just x=1(2.8)
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Old 10-15-2006, 08:30 PM   #3
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i dont think id ever be this bored
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Old 10-15-2006, 08:30 PM   #4
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Man I had this last year and totally forget how to do it.

Sorry, I guess that wasn't much help
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Old 10-15-2006, 08:32 PM   #5
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not too sure how there would be 2 angles
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Old 10-15-2006, 08:41 PM   #6
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Quote:
Originally Posted by azzy989 View Post
hmm for 1 a) you can use x=1/2 at^2...soo its x=-4.9(2.8)^2.
for 1b) you can use x=vavgt so thats just x=1(2.8)
ahhhh thank you, i was missing the ^2 in the first equation somehow..

here's another one i've tried at but can't seem to get..

- A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. The plane is traveling horizontally with a speed of 300 km/h (83.3 m/s).



(a) How far in advance of the recipients (horizontal distance) must the goods be dropped?
_______m

(b) Suppose, instead, that the plane releases the supplies a horizontal distance of 425 m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climbers' position?
Magnitude:
____m/s

(c) With what speed do the supplies land in the latter case?
________m/s
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Old 10-15-2006, 08:43 PM   #7
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Quote:
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not too sure how there would be 2 angles
since it says '2.0m away', i think that just basically means ahead of and behind. so the two angles would be +/- 15 degrees, being 15 and 345 deg. or whatever.
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Old 10-15-2006, 08:48 PM   #8
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hmm as for part a) you do -235=-4.9t^2 then you divide by -4.9 and sqrt to get the time... then you use x=Vavgt so x=83.3(whatever time you got from the first eq)
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Old 10-15-2006, 08:53 PM   #9
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hmm as for part a) you do -235=-4.9t^2 then you divide by -4.9 and sqrt to get the time... then you use x=Vavgt so x=83.3(whatever time you got from the first eq)
Can I fly you out for my physics midterm comin up?
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Old 10-15-2006, 08:53 PM   #10
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whoaa b) looks like a b!tch... well as far as i can see you can figure out time by doing 425= 83.3t .
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Old 10-15-2006, 09:05 PM   #11
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OK got it... -235=Vo(time from just now)+(-4.9)(time from just now)^2

and for c... then you use a=(Vf-Vo)/t so -9.8=Vf-(Vo from b)/ time from a
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Old 10-15-2006, 09:06 PM   #12
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Old 10-15-2006, 09:09 PM   #13
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lol cowmoo, is that webassign.com?
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Old 10-15-2006, 09:11 PM   #14
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Quote:
Originally Posted by Coldintake View Post
Can I fly you out for my physics midterm comin up?
hahaahh sure sure set a time and date jk jk
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Old 10-15-2006, 09:11 PM   #15
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Quote:
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hahaahh sure sure set a time and date jk jk
thanks a billion by the way.. want a photoshop or sig or something?
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Old 10-15-2006, 09:12 PM   #16
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Quote:
Originally Posted by cowmoo32 View Post
If his stuff is too easy, take a shot at mine.....please.....

LOL SORRY haahah we haven't gotten THAT far in physics here
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Old 10-15-2006, 09:14 PM   #17
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Quote:
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thanks a billion by the way.. want a photoshop or sig or something?
ooo that would be cool i'll pm you later or something
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Old 10-15-2006, 09:17 PM   #18
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haha ok cool..

in the meantime, here are more problems. i understand everything but it's just that i don't know how to use the equations.

18.) An athlete executing a long jump leaves the ground at a 35***176; angle and travels 6.80 m.

(a) What was the takeoff speed?
______m/s
(b) If this speed were increased by just 2.0 percent, how much longer would the jump be?
______m

23.) A hunter aims directly at a target (on the same level) 130 m away.

(a) If the bullet leaves the gun at a speed of 280 m/s, by how much will it miss the target?
_____m
(b) At what angle should the gun be aimed so the target will be hit?
_____***176;
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Old 10-15-2006, 09:19 PM   #19
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the ground at a 35***176
thats what came up when you posted...by any chance is the ***176 the degree symbol?? and i'm kinda lost because theres no initial velocity or something like that
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Old 10-15-2006, 09:25 PM   #20
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Quote:
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the ground at a 35***176
thats what came up when you posted...by any chance is the ***176 the degree symbol??
ahh, yes it is. hehe my bad.
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