To solve this problem, we need to find the time it takes for coin A to reach its maximum height and then fall back down to the height of coin B. First, let's find the time it takes for coin A to reach its maximum height. We can use the formula: v = u + at where v is the final velocity (0 m/s at the maximum height), u is the initial velocity (15 m/s), a is the acceleration due to gravity (-10 m/s^2), and t is the time taken. 0 = 15 - 10t t = 1.5 seconds So it takes 1.5 seconds for coin A to reach its maximum height. Now, let's find the height of coin A at its maximum height. We can use the formula: s = ut + 0.5at^2 where s is the displacement (height), u is the initial velocity (15 m/s), a is the acceleration due to gravity (-10 m/s^2), and t is the time taken (1.5 seconds). s = 15(1.5) + 0.5(-10)(1.5)^2 s = 11.25 meters So coin A reaches a maximum height of 11.25 meters above the top of the tower. Next, let's find the time it takes for coin B to fall from the top of the tower to the height of coin A. We can use the formula: s = ut + 0.5at^2 where s is the displacement (height), u is the initial velocity (0 m/s), a is the acceleration due to gravity (-10 m/s^2), and t is the time taken. 11.25 = 0 + 0.5(-10)t^2 t = 1.5 seconds So it takes 1.5 seconds for coin B to fall to the height of coin A. Finally, we can find the distance below the top of the tower where coin A passes coin B. We can use the formula: s = ut + 0.5at^2 where s is the displacement (distance below the top of the tower), u is the initial velocity (15 m/s), a is the acceleration due to gravity (-10 m/s^2), and t is the time taken (1.5 + 1.5 = 3 seconds). s = 15(3) + 0.5(-10)(3)^2 s = 22.5 meters So coin A passes coin B 22.5 meters below the top of the tower.