Problema Solution
During a volcanic eruption, a piece of volcanic ash was ejected from the volcano with an initial velocity of 368 feet/second. The height, in feet, if the ash projectile is given by the equation:
H=-16t2+368t
Where t=time, in seconds.
We will assume that the volcano has no height.
The graph of this equation will be a parabola
1.) When does the projectile reach its maximum height?
2.) What is its maximum height?
3.) When does the ash projectile return to the ground?
Answer provided by our tutors
H=-16t^2+368t
1) t = -b/(2a), where a = -16, b = 368
t = -368/(2*(-16))
t = 11.5 s
The projectile reaches maximum height after 11.5 seconds.
2) we need to find the maximum of the parabolic function H=-16t^2+368t
H max = (4ac – b^2) / 4a, where a = -15, b = 368, c = 0
H max = (4*(-15)*0 - 368^2)/(4*(-15))
H max = 2,257.07 ft
the maximum height is 2,257.07 feet.
3) -16t^2+368t = 0
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click here to see the equation solved for t
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t = 23 s
The projectile returns to the ground after 23 seconds.