A rifle loses 1/20^th of its velocity in passing through a plank.The least number of such planks required just to stop the bullet are

Let us suppose that thickness of each planck be taken as ‘t’.Also suppose ‘n’ no. of plancks are required to stop the bullet.
uniform retardation = a
Initial velocity of rifle = u
Now,after piercing in the first planck,final vaelocity will be
u – u/20 = 19u
20
We know that,
v² = u² + 2as
or,
v² – u² = 2at
(19u/20)² – u² = 2at
or, 2at = – 39u²/400 --------- eq.(1)
When rifle passes through such ‘n’ no. of plancks its final velocity will be 0.
0 – (19u/20)² = 2ant ------------ eq.(2)
substitute the value of 2at from eq.1 in eq.2 we get,
0 – [19u/20]² = n* – 39u²/400

or, n = – 400/--39u²[19u/20 ]²

n = 9.25 or it is taken as 10

therefore,total no. of such plancks will be 10 + 1 = 11