Question

Two particles begin to fall freely from same height but the second falls t seconds after the first.find time ( after dropping of the first) when separation between particles is h

Answer 1

Let after x second the separation between particles is h.
for first particle,
$h_1=ux+\frac{1}{2}gx^2$

$h_1=\frac{1}{2}gx^2$..........(1)

2nd particle,
$h_2=0.t+\frac{1}{2}g(x-t)^2$.......(2)

given, $h_1-h_2=l$

1/2gx² - 1/2g(x - t)² = l

1/2g[x² - (x - t)²] = l

[x² - x² + 2xt - t²] = 2l/g

2xt - t² = 2l/g

2xt = t² + 2l/g

x = t/2 + l/gt