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 \( =8 \) seconds (b) After how many seconds will the ball pass the top of the building on its way down? t \( =\square \) seconds

To determine when the ball strikes the ground, you can set \( s = 0 \) (the height from the ground) and solve the equation: \[ 0 = 128 + 112t - 16t^2 \] Rearranging gives: \[ 16t^2 - 112t - 128 = 0 \] Using the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 16 \), \( b = -112 \), and \( c = -128 \), you find \( t = 8 \) seconds is when the ball strikes the ground. For when the ball passes the top of the building again on its way down, you need to find the time after it has reached its peak. The peak time can be calculated by: \[ t_{\text{peak}} = \frac{v_0}{g} = \frac{112}{32} = 3.5 \, \text{seconds} \] The ball will fall back past the top of the building after another \( 3.5 \) seconds, giving you \( 3.5 + 3.5 = 7 \) seconds. So for part (b), \( t \approx 7 \) seconds. Now, here are some fun facts about your projectile problem: When Galileo first studied projectile motion in the 16th century, he was fascinated by how objects move, influencing modern physics! He conducted experiments using inclined planes and dropped objects which led to the understanding of gravity we have today. In real life, when you throw a ball, especially at a high speed, factors like air resistance and wind can alter its path. Professional athletes often account for these variables during their training sessions to improve their performance. Understanding physics can give you an edge in sports!