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Suppose that a ball is thrown straight up into the air and its height after t seconds is 5 + 32t – 16t^2 feet. Determine the maximum height of the ball.
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well the easy way to do this would be to plug in the equation in a graphing calculator
the time of the maximum height is given by -b/ 2a in this prob b =32 and a = -16. find this value for t and plug it into the quadratic and you got your max height
differentiate to find velocity and we know that when height is max v = 0 so find time then replug into the height equation gives you time is 1 when max and height is 21
Okay, I keep getting the wrong answer, what do you have?
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