During a recent eruption, the volcano spewed out copious amount of ash. one small piece of ash was ejected from the volcano with initial velocity of 368 ft/sec. the height h, in feet, of our projectile is given by the equation h=-16t^2+368t. where t is the time in seconds. we assume that the volcano has no height 1. when does the ash projectile reach its maxim eight? 2. what is its maxim height? 3. when does the ash projectile return to the ground?

For question number 1:The plot H = H(t) is the parabola and it reaches its maximum in the moment when exactly at midpoint between the roots t = 0 and t = 23. At that moment t = 23/2 or 11.5 seconds.
For question number 2:To find the maximal height, just simply substitute t = 11.5 into the quadratic equation. The answer would be 22.9.
For question number 3:H(t) = 0,   or, which is the same as -16t^2 + 368t = 0.Factor the left side to get  -16*t*(t - 23) = 0.t = 0, relates to the very start of the process, when the ash started its way up.The other root is t = 23 seconds, and it is precisely the time moment when the bit of ash will go back to the ground.

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