Hello,

The displacement of a body is given by s = ut + 1/2 * a * t^2

For a freely falling body, u = 0

Therefore, s1 = 1/2 * a * t^2 = 1/2 * 10 * t^2 = 5 * t^2

Therefore, s1 = 5 * t ^2

For a body falling with an initial velocity of 2 m/s, displacement after time t second will be, s2:

s2 = 2 * t + 1/2 *10*t^2

s2 = 2t + 5 * t^2

Now, when the separation between them is 18 m ( I have assumed 18 m in place of 18 years because I think you have typed it incorrect)

Therefore, s2 - s1 = 18

2t + 5 * t^2 - 5 * t^2 = 18

2t = 18

t = 9 seconds (Answer)

I guess the separation between them is 18 metres instead of 18 years. If it is 18 metres then answer of the question would be:-

let the required time be t.

For freely falling body we have,

S1=1/2gt²

=1/2×10×t²

=5t²

For body thrown downward with1m/s,

S2=ut+1/2gt²

=1×t+1/2×10×t²

=t+5t²

Now,S2-S1=18

t=18s

Hope it helps.

Good luck.