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Question: A ball is thrown in the air. The relationship between the amount of time the ball is in the air in seconds (t) and the ball’s distance above the ground in feet (h) is represented by the equation below. h=-16t^2+30t+4 How many seconds will it take for the ball to hit the ground? A.1/8 B.2 C.1/4 D.4

To determine the time it takes for the ball to hit the ground, we need to find when the height \( h \) of the ball is zero. This happens when the ball meets the ground. The given equation is: \[ h = -16t^2 + 30t + 4 \] We set \( h \) to 0 and solve for \( t \): \[ 0 = -16t^2 + 30t + 4 \] This is a quadratic equation of the form \( at^2 + bt + c = 0 \), where \( a = -16 \), \( b = 30 \), and \( c = 4 \). To solve for \( t \), we can use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting in the values of \( a \), \( b \), and \( c \): \[ t = \frac{-30 \pm \sqrt{30^2 - 4(-16)(4)}}{2(-16)} \] \[ t = \frac{-30 \pm \sqrt{900 + 256}}{-32} \] \[ t = \frac{-30 \pm \sqrt{1156}}{-32} \] \[ t = \frac{-30 \pm 34}{-32} \] This gives us two possible solutions: \[ t = \frac{-30 + 34}{-32} = \frac{4}{-32} = -\frac{1}{8} \] \[ t = \frac{-30 - 34}{-32} = \frac{-64}{-32} = 2 \] Time cannot be negative, so we discard \( t = -\frac{1}{8} \). Thus, the ball will hit the ground in: \[ t = 2 \text{ seconds} \] So, the correct answer is: B. 2

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