Eyjafjallajokull Is A Volcano In Iceland During A Recent Eru

Eyjafjallajokull Is A Volcano In Iceland During A Recent Eruption Th

Eyjafjallajokull is a volcano in Iceland. During a recent eruption, the volcano spewed out 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 the ash projectile is given by the equation: H = -16t² + 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 at the beginning, i.e., H=0 at t=0.

Question 1: When does the ash projectile reach its maximum height?

Question 2: What is its maximum height?

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

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The motion of the ash projectile ejected from the volcano can be accurately modeled using the quadratic function H = -16t² + 368t, which represents the height (in feet) as a function of time (in seconds). Variations in astrophysical phenomena such as volcanic ash plumes can be described mathematically using principles of physics and algebra, providing insights into their behavior during eruptions.

Question 1: When does the ash reach its maximum height?

In the quadratic function H(t) = -16t² + 368t, the parabola opens downward because the coefficient of t² is negative, indicating a maximum point at the vertex of the parabola. The time at which maximum height occurs can be found using the vertex formula for a parabola in the form H(t) = at² + bt + c, where t = -b/(2a).

Here, a = -16 and b = 368, so:

t = -368 / (2 * -16) = -368 / -32 = 11.5 seconds.

Thus, the ash reaches its maximum height at t = 11.5 seconds.

Question 2: What is its maximum height?

To find the maximum height, substitute t = 11.5 seconds into the height equation:

H(11.5) = -16(11.5)² + 368(11.5).

Calculate (11.5)² = 132.25, then:

H(11.5) = -16 132.25 + 368 11.5.

H(11.5) = -2116 + 4222 = 2106 feet.

Therefore, the maximum height of the ash projectile is 2106 feet.

Question 3: When does the ash return to the ground?

The projectile hits the ground when H = 0. Setting the height equation to zero:

0 = -16t² + 368t.

Factor out t:

t(-16t + 368) = 0.

This yields two solutions: t = 0 and t = 368 / 16 = 23 seconds.

t = 0 seconds corresponds to the moment of ejection, and t = 23 seconds indicates when the ash hits the ground again.

Hence, the ash returns to the ground after 23 seconds.

References

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