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26 September, 13:17

During a recent eruption, a 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

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?

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Answers (1)
  1. 26 September, 14:06
    0
    The formula is:

    H = - 16 t² + 368 t

    The projectile reaches its maximum at:

    t max = - b / 2 a = - 368 / ( - 32) = 11.5 s

    - 16 t² + 368 t = 0

    t ( - 16 t + 368) = 0

    t 1 = 0 s, t 2 = 368 : 16 = 23 s

    H max = - 16 · 11.5 + 368 · 11.5 = - 2116 + 4232 = 2116 ft

    Answers:

    1. The projectile reaches its maximum height at t = 11.5 s.

    2. H max = 2116 ft.

    3. The projectile returns to the ground after t = 23 s.
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