Ball is thrown upward from the ground with an initial speed of 25 m/s; at the same instant, another ball is dropped from a building 15 m high. after how long will the balls be at the same height?

0.6 seconds after the balls start moving, they'll be at the same height. The altitude the 1st ball will be at time T is a = 25T - 0.5AT^2 The altitude the 2nd ball will be at time T is a = 15 - 0.5AT^2 Set both expressions equal to each other and solve for T 25T - 0.5AT^2 = 15 - 0.5AT^2 Add 0.5AT^2 to both sides 25T = 15 Divide both sides by 25 T = 15/25 = 3/5 = 0.6 So both balls will be at the same altitude 0.6 seconds after they start moving. Let's verify those results. Assume A = 9.8 m/s^2 0.5 * 9.8 m/s^2 * 0.6s^2 = 4.9 m/s^2 *.36 s^2 = 1.764 m The altitude the 1st ball will be at after 0.6 seconds is 0.6 s * 25 m/s - 1.764m = 15 m - 1.764 m The altitude the 2nd ball will be at after 0.6 seconds is 15 m - 1.764 m Both are the same value so the answer is verified.