A particle is dropped under gravity frm rest frm a height h and it travels a distance 9h/25 in the last second. The height is ---

1) 10Q) 0 m        2) 122.5 m       3) 145 m      4) 168.5 m

Let us assume that the particle takes time ' t ' sec. after being dropped from rest from height ' h ' to reach the ground.

As it covers 9h/25 in last second,

so the particle covers (h - 9h/25) = 16h/25 in ' t - 1 ' sec.

Now, using equations of motion

16h/25 = (1/2)g( t - 1)²  ------ (1)

Also, h = (1/2)g( t )² ------- (2)

Solving (1) and (2),

(16/25)t² = (t - 1)²

or, 4t = + (5t - 5)

or, t = 5 sec. ( t = 5/9 is rejected because in question it is given that t>1 as the bosy covers 9h/25 in last second)

So, h = (1/2)g (5)² = (25)(9.8) / 2 = 122.5 m (Ans)..................