water rocket launching upward with velocity of 48 feet/second after how many seconds will the height of the rocket be 20 feet. h=48t-16t^2

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I used a graphing calculator and entered:- 16 t ^2 + 48 tt was replaced by the "x" in the Y = -16x^2 + 48xThen I looked at the table of values generated by the equation.The increments of the "x" values need to be set to .5 or 1/2.This now shows the "x" values going up or down by .5.Ex: 0, .5, 1, 1.5, 2 etc...Look at the values of "y" and there are two places where the rocket reaches a height of 20 feet.  Once on the way up at .5 seconds and once again at 2.5 on the way down.