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Halp!!!!

Name: Anonymous 2007-06-06 8:21 ID:Kd97yUN2

With what velocity must an object be thrown upward from the ground to reach a maximum height of 550 feet?

Name: Anonymous 2007-06-06 8:25 ID:vmTbu4r5

300kb/s

Name: Anonymous 2007-06-06 9:33 ID:XNkpf4bC

>>1
IMPERIALIST

Name: Anonymous 2007-06-06 9:34 ID:XNkpf4bC

>>1
p.s. this is high school level physics, retard

Name: Anonymous 2007-06-06 13:23 ID:EuCpy1Qy

1.21 jiggawatts

Name: Anonymous 2007-06-06 14:23 ID:xhDx8AXp

v^2 = u^2+2as

Name: 4tran 2007-06-06 21:38 ID:Heaven

(v^2)/2 = gh (conservation of energy, divided out common factor of mass)

Name: Anonymous 2007-06-08 17:14 ID:XmnBod2Q

what is the mass of the object? idiot.

Name: Anonymous 2007-06-08 18:50 ID:Heaven

>>8
ahahaha
oh wow

Name: Anonymous 2007-06-08 19:51 ID:poq10ECx

just strap a rocket to it - guaranteed 1000ft

Name: 4tran 2007-06-08 20:00 ID:rjzhrTzL

>>8
We're assuming an object whose mass << that of the earth.  Therefore, we can neglect gravitational perturbations on earth's location.

Name: Anonymous 2007-06-08 22:44 ID:TNYjJklI

Because things that go up generally go down, a parabola will be helpful for you to figure this out.

Name: Anonymous 2007-06-09 18:32 ID:mcTKUawi

idiots, all of you. except >>6 and >>7, who already gave the answer.

The formula you want is v^2 = u^2 + 2as. u=0, a=g, s=550 feet=167.64 meters (stupid fucking imperial system)

v = sqrt(2*9.8m/s*167.64m) = 57.32 m/s = 128 mph

Name: Anonymous 2007-06-10 3:21 ID:YeZLTuUQ

>>13
>>13
Yes, on the assumption that the object is of negligible mass compared to the earth, the absence of air resistance and wind, and the change in gravitational pull is negligible at 550feet as compared to that of sea level.

Name: Anonymous 2007-06-10 8:05 ID:aTWJUBu9

>>13

I'm gonna have to say no. u is not 0, u is what you want to find. You're trying to find the initial speed vertically upwards to travel s=167.64. v may be 0, but we don't know that, all we know is that v >= 0 at s=167.64. Also, you had a=g and then s > 0, when in fact if a > 0 then you need to stipulate that s < 0, since you're working in the downward direction. What you ACTUALLY wanted was a = -g UPWARDS.

Where do you morons come from?

Name: Anonymous 2007-06-10 10:08 ID:YeZLTuUQ

>>15
lulz.

v^2=u^2+2as.
Taking direction upward as positive,
s = +167.64m
a = -9.81m/s^2
v = 0m/s
Problem solved.

v of course will be zero since it is at maximum height. Suppose v is positive it means it'll go higher than 167.64 meters which means 167.64 meters is not its maximum height, which means you aren't answering the question.

Name: 4tran 2007-06-11 4:05 ID:mgIaaAWx

You could also consider the object at maximum height falling (from rest) the same distance.  How fast it hits the ground is the answer you want.

With down as the + direction, >>13's answer is still right.

Name: Anonymous 2009-03-18 3:05

Don't call me gay, but I need some mary jay!

Marijuana MUST be legalized.

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