KCET 2025 Physics Paper

To solve this problem, we need to determine the time, denoted as t0, at which the separation between two falling stones becomes H . The second stone starts falling ∆t seconds after the first stone.
Distance Fallen by the First Stone ( S1 ):
The first stone falls for t0 seconds, so the distance it covers is given by:
S1=‌

1

2

gt02
Distance Fallen by the Second Stone (S2) :
The second stone starts falling ∆t seconds after the first, so it falls for (t0−∆t) seconds. Thus, the distance it covers is:
S2=‌

1

2

g(t0−∆t)2
Distance of Separation (H):
The separation distance between the two stones is:
H=S1−S2=‌

1

2

gt02−‌

1

2

g(t0−∆t)2
Simplifying the equation:
H‌=‌

1

2

gt02−‌

1

2

g(t02−2t0∆t+(∆t)2)
‌=‌

1

2

g⋅(2t0∆t−(∆t)2)
‌=g⋅(t0∆t−‌

1

2

∆t2)
Expression for t0 :
Solving the above equation for t0, we have:
‌H=g(t0∆t−‌

1

2

∆t2)
‌t0∆t=‌

H

g

+‌

1

2

∆t2
‌t0=‌

H

g∆t

+‌

∆t

2

Therefore, the time t0 at which the separation between the two stones becomes H is given by:
t0=‌

H

g∆t

+‌

∆t

2