Yahoo Answers: Answers and Comments for You throw a stone straight down. the stone starts with a velocity of 54 feet per second down. it leaves the sling 326 feet above the ground? [Physics]
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From Josie
enUS
Sun, 19 May 2019 15:27:13 +0000
3
Yahoo Answers: Answers and Comments for You throw a stone straight down. the stone starts with a velocity of 54 feet per second down. it leaves the sling 326 feet above the ground? [Physics]
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https://answers.yahoo.com/question/index?qid=20190519152713AAlD1u8
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From Andrew Smith: Adding a question mark does not make a questio...
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Sun, 19 May 2019 17:36:56 +0000
Adding a question mark does not make a question.
ASKING a question makes a question.
What did you hope to find?
AN equation could be as simple as
h = 326
or
v = 54
These are both equations.
as is h = vt + 1/2 g t^2
or v^2 = u^2 + 2gh
v = u + gt
All of these are equations that MIGHT be relevant.
And I could think of so many more.
To ask a question you need to have some clear idea of what it is that you need to find.

From electron1: There are two numbers that you can find. They ...
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Sun, 19 May 2019 16:59:24 +0000
There are two numbers that you can find. They are the stone’s velocity just before it hits the ground and time the stone is moving. The acceleration is 32 ft/s^2. Use the following equation to determine the stone’s velocity just before it hits the ground.
vf^2 = vi^2 + 2 * a * d
vf^2 = 54^2 + 2 * 32 * 326
vf = √23,780
This is approximately 154 ft/s. Use the following equation to determine the time.
vf = vi + a * t
√23,780 = 54 + 32 * t
t = (√23,780 – 54) ÷ 32
This is approximately 3.13 seconds. If you use a slightly different number for the acceleration, your answers will be slightly different. I hope this is helpful for you.

From Lôn: It depends on what you are trying to find...i)...
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Sun, 19 May 2019 16:16:04 +0000
It depends on what you are trying to find...i)the final velocity (v) or ii) the time it takes to reach the ground (t)
i) v^2 = u^2 + 2aS
ii) S = ut + at^2/2

From oldprof: The stone's height above impact point is g...
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Sun, 19 May 2019 15:51:48 +0000
The stone's height above impact point is given by y(t) = h  Ut  1/2 gt^2; where h = 326 ft, U = 54 fps, and g = 32.2 ft/s^2. To solve for t = ? time to impact, we set y(t) = 0 and solve the quadratic eqn.

From 冷眼旁觀: The question is incomplete. Please check.
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Sun, 19 May 2019 15:39:52 +0000
The question is incomplete. Please check.