To find out when the ball will hit the ground, we need to determine when the height \( h \) is equal to zero. So we set the equation \( h = -16t^2 + 30t + 4 \) equal to zero and solve for \( t \): \[

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