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Question

# A body is dropped from the top of a tower. During the last second of its fall, it covers 16/25th of the height of the tower. Calculate the height of the tower.

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Solution

## Step 1: Given Let the height of the tower be ‘h’. Suppose it takes ‘n’ second to reach the ground. 'Sn' is height covered in (n−1) second.Step 2: Formula usedS=ut+a2(t)2 Step 3: Finding the time required to cover the height of the towerIn nth second the body covers a height of 1625h.In (n-1)th second, the body covers a height of (1-1625h) = 925h It starts from rest, thus, Sn=u+a2(n−1)2 [this is the equation to find the distance traveled in (n-1)th second] = (9h/25) = 0 + (9.8/2)(n – 1)² = 9h/25 = 4.9(n – 1)² …...(1) Again, the body reaches the ground in ‘n’ second. So, the distance traveled in ‘n’ second is, h = 0 + ½ gn2 => h = (9.8/2)n2 => h = 4.9n2 …………….…(2) From (1) and (2) we have, 9×4.9×n225 = 4.9(n – 1)² => 9n2 = 25(n – 1)² => 16n2 – 50n + 25 = 0 => n = 0.625 or 2.5 Here, n = 0.625 s is not possible as it is greater than 1. So, the time in which the body hits the ground is 2.5 s.Step 4: Finding the height of the tower Putting value of n in equation (2), we geth=4.9n2=4.9×2.5×2.5=30.625h = 4.9n^2=4.9\times 2.5\times 2.5=30.625 m. Hence, the height of the tower is 30.625 m.

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