a man drops a ball downside from the roof of a tower of height 400 meters. At the same time another ball is thrown upside with a velocity 50 m/s from the surface of the tower, find when and at which height from the surface of the tower the two balls meet together.

Case A: Ball dropped downwards initial vel., v0=0m/s accl, a=-g=-9.8m/s^2 displacement, S = (x-400) m Then we have S=v0*t+0.5a*t^2 x-400 = - 0.5*9.8*t^2Case B: Ball thrown upwards initial vel., v0=50m/s accl, a=-g=-9.8m/s^2 displacement, S = (x-400) m Then we have S=v0*t+0.5a*t^2 x-400 = 50*t-0.5*9.8*t^2We have 2 eqns: (x-400) = - 0.5∗9.8∗ t2 (x-400) = 50∗t-0.5∗9.8∗ t2 Solving, we get t=0 & x=400 This is the initial state ... which means that they never meet during the flight ...