A bullet is fired with a speed of 450m/s through a board 20 cm thick. If the bullet is subject to a large deceleration of magnitude 106m/s2, will the bullet be able to emerge from the other side of the board ?

Answer 1

The distance X required to stop the bullet when decelerating from Vo = 450 m/s at that rate (a) is given by

Vo = sqrt(2 a X), which can be rewritten

X = Vo^2/(2a)

It is not clear whether your deceleration rate is 106 m/s^2 or 10^6 m/s^2. It makes a difference in your answer.