A body falling freely from rest covers 7/16 of total height in the last second of its fall. The height from which it falls is? (g = 9.8 m/s^2)?

A body falling freely from rest covers 7/16 of total height in the las

1.	A body falling freely from rest covers 7/16 of total height in the last second of its fall. The height from which it falls is? (g = 9.8 m/s^2)?
Answer» $s = ut + \frac{1}{2} a {t}^{2}$ $\frac{9}{16} = 0 + \frac{1}{2} 9.8 {t}^{2} \\ \frac{18}{16} = 9.8 {t}^{2} \\ \frac{1.83}{16} = {t}^{2} \\ t = \frac{ \sqrt{1.83} }{ \sqrt{16} } \\ t = \frac{1.35}{4}$ hope this will HELP you. further QUERY ask in COMMENT SECTION

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A body falling freely from rest covers 7/16 of total height in the last second of its fall. The height from which it falls is? (g = 9.8 m/s^2)?

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