A projectile is fired straight up from ground level with an initial velocity of 112 ft/s. Its height, h, above the ground after t seconds is given by h = –16t2 + 112t. What is the interval of time during which the projectile's height exceeds 192 feet?

Set the height greater than 192: −16t² + 112t > 192. Rearranging and dividing by −16 reverses the inequality, giving t² − 7t + 12 < 0, or (t − 3)(t − 4) < 0. This expression is negative between its roots, so the projectile is above 192 feet when 3 < t < 4 seconds, choice A.

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