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=v0t+0.5at^2
x-400= -0.59.8t^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=v0t+0.5at^2
x-400= 50t-0.59.8*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...
