Answer
The two diamonds will be ten meters apart $~~1.52~seconds~~$ after the first diamond begins to fall.
Work Step by Step
We can write an expression for the distance the first diamond falls:
$y_1 = \frac{1}{2}at^2$
$y_1 = \frac{1}{2}(9.8)t^2$
$y_1 = 4.9~t^2$
We can write an expression for the distance the second diamond falls:
$y_2 = \frac{1}{2}a(t-1)^2,~~~~~t \geq 1.0~s$
$y_2 = \frac{1}{2}(9.8)(t^2-2t+1)$
$y_2 = (4.9)(t^2-2t+1)$
$y_2 = 4.9t^2-9.8~t+4.9$
We can find $t$ when $y_1 = y_2+10$:
$y_2+10 = y_1$
$(4.9t^2-9.8~t+4.9)+10 = 4.9~t^2$
$4.9t^2-9.8~t+14.9 = 4.9~t^2$
$-9.8~t+14.9 = 0$
$9.8~t=14.9$
$t = \frac{14.9}{9.8}$
$t = 1.52~s$
The two diamonds will be ten meters apart $~~1.52~seconds~~$ after the first diamond begins to fall.
