Question

A bullet loses 1/20th of its velocity when passing through a plank. what is the least number of planks required to stop the bullet?

Answer 1

If the bullet looses 1/20th of its velocity then

V= 19/20u

Since, v² = u² + 2as

(19/20u)² = u² - 2as

2as = u² - (19/20u)² ................(1)

Now, initial velocity = 0

Therefore 0² = u² - 2a (ns)

u² = 2as (n)

u² = n [ u² - 19²/20² u²]

u² = nu² [ 20²-19²/20²]

20² = n [ 20² - 19² ]

20² = n [ (20+19)(20-19) ]

n = 400/39

n ~ 11 planks


hope it will help you..

Answer 2

Answer:

11 planks

Step-by-step explanation:

let the initial velocity of the bullet be u

then......(in the attachment)