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a bullets hits a sand box with a velocity of 20m/s and it penetrates its up to a distance of 6 cm. find the decelaration of the bulletvin the sand box |
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Answer» Explanation:Solution,Here, we haveInitial velocity of bullets, u = 20 m/sFinal velocity of bullets, v = 0Distance COVERED, C = 6 cm = 0.06 mTo Find,Deceleration of the bullet, a = ?According to the 3rd law of motion,We know that,v² - u² = 2asSo, PUTTING all the values, we get⇒ v² - u² = 2as⇒ (0)² - (20)² = 2 × a × 0.06⇒ 400 = 0.12a⇒ 400/0.12 = a⇒ a = 3333.33 m/s²Hence, the deceleration of the bullet is 3333.33 m/s². |
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