A stone is dropped from the top of a tower
of height h=60 m. Simultaneously another
stone is projected vertically upwards from
the foot of the tower. They meet at a height
2h
from the ground level. The initial
velocity of the stone projected upwards is
(g = 10 ms?).
[AP E-2017]
please give an explanation
Answers
A stone is dropped from a height h.simultaneously another stone is thrown up from the ground which reaches the height 4h.the two stones cross each other after a time-------------
7 years ago
Answers : (2)
For upper stone which is dropped downwards the distance traveled is
S1 = ut + 1/2 g t^2
Here u = 0
g = accln due to gravity = 9.8 m/s2
So S1 = 1/2g t2
Distance of stone from ground is = h - S1 = h - 0.5 g t2 ....(1)
Now for stone thrown upwards such that it goes to height 4h from ground
Potential energy at top = Kinetic energy from the point of projection
or mg(4h) = 1/2 m u2
or u = (8hg)1/2
Thus S2 = ut - 0.5 g t2
or S2 = (8hg)1/2 t - 0.5 g t2 ........(2)
Now for the two stones to meet
h - S1 = S2
or h - 0.5 g t2 = (8hg)1/2 t - 0.5 g t2 (from 1 and 2)
or h = (8hg)1/2 t
or t = ( h/8g) 1/2