Height of a tower =125m The distance sn covered in the nth second is given by sn=ut+‌
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g(2n−1)‌‌‌⋅⋅⋅⋅⋅⋅⋅(i) where u is the initial velocity, which is zero for a body falling from rest. Using the 2 nd equation of motion, s=ut+‌
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gt2 125=‌
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×10×t2 T=√‌
2×125
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=5s For Eq. (i), s5‌=0+‌
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×10(2×5−1) ‌=‌
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×10(10−1)=‌
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×10(9) ‌=5×9 s5‌=45m‌‌...‌ (ii) ‌ The percentage of the height of the tower at this distance, ‌x=‌
