A ball is thrown vertically upward from the top of a building 128 feet tall with an initial velocity of 112 feet per second. The distance \( s \) (in feet) of the ball from the ground after \( t \) seconds is \( s=128+112 t-16 t^{2} \) (a) After how many seconds does the ball strike the ground? (b) After how many seconds will the ball pass the top of the building on its way down? (a) After how many seconds does the ball strike the ground? \( t=\square \) seconds
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To determine when the ball strikes the ground, we need to set the distance function \( s \) equal to zero (since the ground level is defined as 0 feet): \[ s = 128 + 112t - 16t^2 = 0 \] Rearranging gives us the equation: \[ -16t^2 + 112t + 128 = 0 \] To find \( t \), we can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -16 \), \( b = 112 \), and \( c = 128 \). First, let's calculate the discriminant: \[ b^2 - 4ac = 112^2 - 4(-16)(128) = 12544 + 8192 = 20736 \] Now apply the quadratic formula: \[ t = \frac{-112 \pm \sqrt{20736}}{2 \cdot -16} = \frac{-112 \pm 144}{-32} \] Calculating both potential solutions gives: 1. \( t = \frac{-112 + 144}{-32} = \frac{32}{-32} = -1 \) (not physical since time cannot be negative) 2. \( t = \frac{-112 - 144}{-32} = \frac{-256}{-32} = 8 \) So, the ball strikes the ground after \( t = 8 \) seconds. **Answer:** (a) After how many seconds does the ball strike the ground? \( t = 8 \) seconds (b) To find the time when the ball passes the top of the building on its way down, we check when the ball reaches 128 feet again. We set: \[ 128 + 112t - 16t^2 = 128 \] Which simplifies to: \[ 112t - 16t^2 = 0 \] Factoring gives: \[ 16t(7 - t) = 0 \] This yields two solutions: 1. \( t = 0 \) (the initial launch) 2. \( t = 7 \) The ball passes the top of the building on its way down after \( t = 7 \) seconds. **Answer:** (b) After how many seconds will the ball pass the top of the building on its way down? \( t = 7 \) seconds