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A bullet with a velocity of 40 m/s strikes a Sandbox and penetrates it up to a distance of 8 m. Find the bullet's deceleration in the sandbox.

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250 ms^-2

10000 ms^-2

30000 ms^-2

2.5 ms^-2

answer is C.

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Detailed Solution

Distance = 8 cm = 0.08 m

v = 40 m/s

u = 0 m/s

v²=u² +2as

Here u = 0

Then v²= 2as

40 x 40 = 2 x a x 0.08

a = - 10000 m/s²

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