Arocket is launched vertically from the ground with an initial velocity of 64 ft/s (a) write a quadratic function, h(t), that shows the height, in feet, of the rocket t seconds after it was launched (b) graph h(t) on a coordinate plane (c) use your graph from part b to determine the rockets maximum height, the amount of time it took to reach its maximum height, and the amount of time it was in the air. show how you came to the answer.

First we write the cinematic equation:   h (t) = a * (1/2) * (t ^ 2) + v0 * t + h0  where,  h (t): height after t seconds  a: acceleration  t: time  v0: initial speed  h0: initial height  substituting values we have:   h (t) = - 32 * (1/2) * (t ^ 2) + 64 * t + 0  rewriting:   h (t) = - 16 * (t ^ 2) + 64 * t  we look for the maximum, for this we derive:   h '(t) = - 32 * t + 64  we equal zero and clear t:   0 = -32 * t + 64  32 * t = 64  t = (64) / (32)  t = 2  we look for the second derivative to verify that it is a maximum:     h '' (t) = - 32  evaluating for t = 2  h '' (2) = - 32 (< 0 is a maximum)  the maximum height will then be:   h (2) = - 16 * ((2) ^ 2) + 64 * (2)  h (2) = 64 feet  the amount of time that was in the air is given by:   -16 * (t ^ 2) + 64 * t = 0  resolving:   t * (- 16t + 64) = 0  -16t + 64 = 0  16 * t = 64  t = 64/16  t = 4 s  answer:   the quadratic function is:   h (t) = - 16 * (t ^ 2) + 64 * t  the maximum height is:   64 feet  the time to reach this height is:   t = 2 s  the amount of time it was in the air is:   t = 4 s

Logical equivalent of the give question is its contrapositive