Question

A stone is dropped from the top of a tower of height h = 60m. Simultaneously another
stone is projected vertically upwards from the foot of the tower. They meet at a height
of 2h/3 from the ground level. The initial velocity of the stone projected upwards is --
-- msec-1

Answer 1

Answer:

30m/s

Given height of tower=60m

given from ground level it reach =2h/3

so from top it reaches= h - 2h/3= h/3

from top 60/3=20m

s=ut+1/2at2

20=0+1/2×10×t2 (as it is from rest intital velocity u=0)

t=2secs

both meet at 2h/3 means it also take 2secs to reach 2h/3 distance

2h/3=40m

40 = 2u - 0.5*10*2*2

60/ 2 = 30 m/s = u

u = 30 m/s