Question

A bullet hits a wall with a velocity of 20 m/sec and panetrates it up to a
distance of 5 cm. Find the decceleration of the bullet in the wall.​

Answer 1

v^2= 2aS here S is 5 cm(0.05m), and v is 20m/sec. a is deacceleration , 20×20=2a×0.05 after solving it a is 4000 m/sec^2

Answer 2

Answer:

U= 20 m/s

v= 0

s = 5 cm = 0.05 m

from 3rd equation,

v² -u² = 2as

⇒0 - (20)² = 2×0.05×a

⇒0.1 a = - 400

⇒a = - 4000 m/s²

deceleration will be - 4000 m/s²

HOPE IT HELPS YOU

PLEASE MARK ME AS THE BRAINLIEST