Question

a rock is thrown from the top of a tall building. The distance, in feet, between the rock and the ground t seconds after it is thrown is given by d(t)= -3t^2 + 2t + 24. how long after the rock is thrown it is 16 feet from the ground?

Answer 1

• After 2 seconds

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We need to solve the equation for t when d(t) is 16:

• -3t² + 2t + 24 = 16
• -3t² + 2t + 8 = 0
• t = (-2 ± √(2²- 4(-3)*8))) / (-6)
• t = -(-2 ± √100) / 6
• t = - (-2 ± 10) / 6
• t = 2 or t = -8/6 (rejected as negative)

After 2 seconds the rock is 16 feet from the ground.