Standing near the edge of the top of a building, a person throws two balls A and B . Ball A is thrown vertically upward and ball B is thrown vertically downward with the same speed. If ball A strikes the ground with a speed v_A and ball B strikes the ground with a speed v_B , then

Step-by-step solution

Let the height of the building be h and the initial speed of both balls be u . For ball A , taking the downward direction as positive, the initial velocity is -u and the displacement is h . Using the equation of motion: v_A^2 = (-u)^2 + 2gh = u^2 + 2gh For ball B , the initial velocity is u downwards and the displacement is h . Using the equation of motion: v_B^2 = u^2 + 2gh From the above equations, v_A^2 = v_B^2 , which implies v_A = v_B . Alternatively, ball A goes up and returns to the point of projection with