Question

A bullet hits a wall with a velocity of 20 m/s and penetrates it up to a distance of 5 cm. Find the deceleration of the bullet in the wall.

Answer 1

Dtep1 Given:

Initial velocity = $20m/s$

Distance = $5cm=0.05m$

Step2 Formula Used:

$v^2=u^2-2as$

Step3 Calculation for deacceleration:

Substituting the values

$0=20\times 20-2\times a\times .05$

$a=\frac{20\times 20}{2\times 0.05}=4000m/s^2$

Hence deceleration is $4000m/s^2$.