A ball is thrown straight upward. Suppose that the height of the ball at time t is given by the formula h = - 18t² + 126t, where h is in feet and t is in seconds, with t = 0 corresponding to the instant that the ball is first tossed. At what time(s) does the ball reach a height of 108 ft? A) 1 sec, 6sec B) 4 sec C) 3 sec D) 8 sec E) 3 sec, 8sec

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Quizplus · Accepted Answer

To find when the ball reaches a height of 108 ft, set the given height equation equal to 108 and solve for t: $108 = -18t^2 + 126t$. Rearranging gives $18t^2 - 126t + 108 = 0$, which simplifies to $t^2 - 7t + 6 = 0$. Factoring this equation gives $(t - 1)(t - 6) = 0$, so $t = 1$ sec and $t = 6$ sec are the times when the ball reaches 108 ft.

A Ball Is Thrown Straight Upward