A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectile is modeled using the equation h(t) = –16t2 + 48t + 190.Approximately when will the projectile hit the ground?

Set the height equal to zero to find when the projectile reaches the ground: −16t² + 48t + 190 = 0. The positive solution is t = (48 + √14,464) ÷ 32 ≈ 5.26 seconds; the negative solution does not represent a future time. So the projectile hits the ground at approximately 5.3 seconds, choice C.

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