a bullet hits a wall with a velocity of 20m/s and penetrates it up to a distance of 5cm. find the deceleration of the bullet in the wall
Answers
Answered by
14
hi friend,
given u=20m/s
→ s=0.05m.
→v=0m/s( when the bullet stops)
we know that v²-u²=2as
→-400=2(0.05)a
→-200=(0.05)a
→a=-4000m/s²
I hope this will help u ;)
given u=20m/s
→ s=0.05m.
→v=0m/s( when the bullet stops)
we know that v²-u²=2as
→-400=2(0.05)a
→-200=(0.05)a
→a=-4000m/s²
I hope this will help u ;)
Answered by
6
v^2=u^2+2as
Initial velocity=20m/s
Final velocity=0m/s
Distance=0.05m
So putting the value we get
0=400+0.1a
-400/0.1=-4000m/s^2 will be dissceleration
Initial velocity=20m/s
Final velocity=0m/s
Distance=0.05m
So putting the value we get
0=400+0.1a
-400/0.1=-4000m/s^2 will be dissceleration
InfinityMS:
Upper sorry I multiplied it by 5 it should be 0.05 sorry by mistake....and I m not getting edit option...so pls check that portion
Similar questions