Question

52. A ball is thrown upward from the ground with an ini- M tial 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 above the ground?

please explain each step and provide the formula needed to solve this problem

Answer 1

Apply kinematric equation to the ball A y = y_0 + 1/2at^2 + V_0t y = 0 + 1/2(- 9.8)t^2 + 25 t - (1) to the ball B: - y = 15 + 1/2(- 9.8)(t^2)+25 t - (2) (1) - (2) then - 4.9t^2 + 25t = - 4.9t^2 + 15 25t = 15 t = 15/25 = 0.6 sec 0.6 sec will the balss be at the same height above the ground