Yahoo Answers: Answers and Comments for Assume the acceleration of the object is
a(t) = −9.8
meters per second per second. (Neglect air resistance.)? [Physics]
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https://answers.yahoo.com/question/index?qid=20190423125730AAFPOhO
From Anonymous
enUS
Tue, 23 Apr 2019 12:57:30 +0000
3
Yahoo Answers: Answers and Comments for Assume the acceleration of the object is
a(t) = −9.8
meters per second per second. (Neglect air resistance.)? [Physics]
292
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https://answers.yahoo.com/question/index?qid=20190423125730AAFPOhO
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From electron1: At the maximum height, the object’s velocity i...
https://answers.yahoo.com/question/index?qid=20190423125730AAFPOhO
https://answers.yahoo.com/question/index?qid=20190423125730AAFPOhO
Tue, 23 Apr 2019 14:45:02 +0000
At the maximum height, the object’s velocity is 0 m/s. Let’s use the following equation to determine its initial velocity.
vf^2 = vi^2 + 2 * a * d
d = final height – initial height = 240 – 3 = 237 meters
0 = vi^2 + 2 * 9.8 * 237
vi = √4,645.2
The initial velocity is approximately 68 m/s.

From oldschool: You could use conservation of energy (neglecti...
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https://answers.yahoo.com/question/index?qid=20190423125730AAFPOhO
Tue, 23 Apr 2019 14:23:23 +0000
You could use conservation of energy (neglecting air resistance)
mg(2403) = ½mVi²
g(2403) = ½Vi²
2*g(2403) = Vi² = 68.2² so the launch velocity is about 68m/s to reach a height of 240m from a 3m launch height.
OR we could do it this way
Height(t) = ho + Vyi*t  ½*g*t²
But Vyi  gt = 0 at max height so t = Vyi/g at h(t) = 240m
240 = 3 + Vyi²/g  ½*g*Vyi²/g² > 237 = Vyi²/2g >
19.6*237 = Vyi² = 68.2m/s same answer.

From Some Body: v² = v₀² + 2a(x  x₀)
(0 m/s)² = v₀² + 2(9.8...
https://answers.yahoo.com/question/index?qid=20190423125730AAFPOhO
https://answers.yahoo.com/question/index?qid=20190423125730AAFPOhO
Tue, 23 Apr 2019 13:04:12 +0000
v² = v₀² + 2a(x  x₀)
(0 m/s)² = v₀² + 2(9.8 m/s²) (240 m  3 m)
v₀ =