Eyjafjallajokull is a volcano in Iceland During a recent eruption, the volcano spewed out copious amounts of ash

Anonymous
timer Asked: Dec 14th, 2017
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Question description

E is a volcano. During a recent eruption, the volcano spewed copious amounts of ash. One small piece of ash was ejected from the volcano with an initial velocity of 368 ft/sec. The height H, in feet, of our ash projectile is given by the equation:

H = -16t^2 + 368t,

Where t is the time in seconds. The graph of this equation will be a parabola. We will assume that the volcano has no height. In other words, when H=0 at t=0.

Question 1. When does the ash projectile reach the its maximum height? t=?

Question 2. What is its maximum height?

Question 3. When does the ash projectile return to the ground?

Tutor Answer

Borys S
School: UC Berkeley

The solution is ready, please ask if something is unclear.docx and pdf files are identical

Note that H(t)=16t(-t+23) because 368=16*23.
1. The maximum of a parabola branches down is reached at t=-b/(2a), where a=-16...

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Anonymous
Goes above and beyond expectations !

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