Question

An object is allowed to fall freely from a tower of height 39.2m. Exactly at the same time another stone is thrown from the bottom of the tower in vertically upward direction with a velocity of 19.6 m/s. Calculate when and where these two stones would meet?

Answer 1

hey here is your answer________

relative displacement=39.2m

relative initial velocity ={0-(19.6)}=19.6m/s

relative acceleration ={(-9.8)-(-9.8)}=0

so S=ut+1/2at^2

S=ut

39.2=19.6×t

so t=2 sec

Answer 2

Answer:

t=2s , x=19.6m

hope it's best explanation