Question

A stone is allowed to fall from the top of a tower 100 m
and at the same time another stone is projected verti
upwards from the ground with a velocity of 25 m/s. Call
when and where the two stones will meet.​

Answer 1

Answer:

for the first stone

height=100-x

u=0

t=?

g=9.8m/s

h=ut+1/2gt^2

100-x=0*t+1/2*9.8.t^2

100-x=4.9t^2

for stone 2

h=xm

u=25

t=?

g,=9.8

s=ut+1/2gt^2

x=25t*4.9t^2

on adding we get

100-x+x=4.9t^2+25t-4.9t^2

100=25t

t=4sec

now putting the value of

100*x=4.9t2

100-x=4.9*16

x=21.6