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Bullet loss half of the its velocity in passing through a plank what is the least number of bank required to stop the bullet Bullet loss half of the its velocity in passing through a plank what is the least number of bank required to stop the bullet
Let a = acceleration/deceleration of the bullet inside the plank.let s = thickness of the planklet the initial velocity of the bullet before hitting the plank: = ufinal velocity = v = u - u/n = u (n-1)/n v² = u² + 2 a s u² (n-1)² / n² = u² + 2 a s 2 a s = u² [ (n-1)² - n² ] / n² = - (2n - 1) u² / n² a = - (2 n - 1) u² / (n² 2 s)let S be the thickness of the N planks kept one after another to stop the bullet. The deceleration is the same. The initial velocity is the same. v² = 0 = u² - 2 S * (2 n - 1) u² / (n² 2 s) 1 = S/s * (2 n -1) / n² N= S/s = n² / (2 n - 1) = 1/2 * [ (2n² -n) + n ] / (2n -1) = 1/2 * [ n + n / (2n -1) ] = n/2 + n/(4n-2) the second term above is between 0 and 1. N = number of planks = [ n/2 ] + 1 where [ n/2 ] is the greatest integer function for n/2. now in this question n = 2 hence total number of plank required = [2/2] + 1 = 1 + 1 = 2
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