# numerical problem

We know that, s = ut + 12(t^2)

Since body starts from rest, u = 0 m/s

y1 = 1/2 a(t^2)

y1 = 1/2 a(10^2)

y1 = 50a——(i)

Now, intial velocity for y2 is the final velocity of y1

Let the final velocity of y1 = initial velocity of y2 = v

So, v = u + at

v = at

v = 10a

Therefore, s2 = ut + 1/2 a(t^2)

s2 = (10a)10 + 1/2 a(10^2)

y2 = 150a——–(ii)

From (i) and (ii),

y1/y2 = 50a/150a

3y1 = y2

Hence, the relation between y1 and y2 is 3y1 = y2