A toy rocket is launched straight up into the air with an initial velocity of 60 ft/s from a table 3 feet above the ground. If acceleration due to gravity is negative 16 the fraction with numerator ft and denominator s squared , approximately how many seconds after the launch will the toy rocket reach the ground? h of t is equal to A t squared plus v t plus h sub 0

Use the height formula with acceleration −16 ft/s², initial velocity 60 ft/s, and starting height 3 ft: h(t) = −16t² + 60t + 3. The rocket reaches the ground when h(t) = 0; solving gives the positive time t = (60 + √3792) / 32 ≈ 3.80 seconds. The negative root represents a time before launch, so the answer is C) 3.80 s.

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