Let's break down the problem step by step. **Given:** - Distance, \( d = 70 \) m - Initial velocity, \( v_0 = 18 \) m/s - Deceleration (negative acceleration), \( a = -1.0 \) m/s² We need to find the time, \( t \), it takes for the ball to reach you. --- ### Step 1: Use the Kinematic Equation The kinematic equation relating distance, initial velocity, acceleration, and time is: \[ d = v_0 t + \frac{1}{2} a t^2 \] Plug in the known values: \[ 70 = 18t + \frac{1}{2}(-1.0)t^2 \] \[ 70 = 18t - 0.5 t^2 \] --- ### Step 2: Rearrange into a Standard Quadratic Equation \[ -0.5 t^2 + 18t - 70 = 0 \] Multiply both sides by -2 to clear decimals: \[ t^2 - 36t + 140 = 0 \] --- ### Step 3: Solve the Quadratic Equation This is a standard quadratic equation: \( t^2 - 36t + 140 = 0 \) Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): - \( a = 1 \) - \( b = -36 \) - \( c = 140 \) \[ t = \frac{-(-36) \pm \sqrt{(-36)^2 - 4 \cdot 1 \cdot 140}}{2 \cdot 1} \] \[ t = \frac{36 \pm \sqrt{1296 - 560}}{2} \] \[ t = \frac{36 \pm \sqrt{736}}{2} \] Calculate \( \sqrt{736} \): \[ \sqrt{736} \approx 27.13 \] So, \[ t = \frac{36 \pm 27.13}{2} \] --- ### Step 4: Find the Physical Solution We need the positive and realistic value for time: 1. \( t_1 = \frac{36 + 27.13}{2} = \frac{63.13}{2} = 31.565 \) s 2. \( t_2 = \frac{36 - 27.13}{2} = \frac{8.87}{2} = 4.435 \) s Since \( t_1 \) is much larger and doesn't make physical sense (the ball would have already stopped), the **reasonable answer is**: --- ## **Final Answer** \[ \boxed{t \approx 4.44 \,\text{seconds}} \] So, it takes **about 4.44 seconds** for the ball to reach you.