Question

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.

please help having test tomorrow and cant find answer to this question

Answer 1

Assume the event happens at time=t and at a distance x from the origin...
Take good care of the axis drawn...

Answer 2

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^2

Answer 3

Case 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^2

Answer 4

We have 2 eqns:
\[(x-400)=-0.5*9.8*t^2 \]
\[(x-400)=50*t-0.5*9.8*t^2 \]
Solving, we get
t=0 & x=400

This is the initial state.... which means that they never meet during the flight...

Answer 5

where did u get the image from?

Answer 6

i drew it... using open office draw...