TS EAMCET 5 May 2018 Shift 1 Solved Paper

The motion of the two particles is represented in the following figure. Let t be the time after the stones meet each other at someheight from the ground.
The distance travelled by first stone in t second is 100−h
And the distance travelled by second stone in t second is h.
Apply kinematic equation for stone I.
−(100−h)=0−

1

2

gt2 …… (i)
Apply kinematic equation for stone I.
h=25t−

1

2

gt2 …… (ii)
On solving the above equation we get,
t=

100

25

=4s