When a body is dropped from a tower, it covers 75% of total height of the tower in the last second of its fall. What is the total height of the tower?

4.9

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9.8

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5.6

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6.0

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Solution

The correct option is B $4.9$ m
Let h be the height of the tower the initial velocity u=0
$h = \frac{1}{2} {g t}^{2} S_{n} = \frac{g}{2} (2 t - 1) = 0.75 h = \frac{g}{2} (2 t - 1) = 0.75 (\frac{1}{2} {g t}^{2}) (f r o m 1)) 2 t - 1 = \frac{75 t^{2}}{100} = \frac{3}{4} t^{2} 8 t - 4 = {3 t}^{2} {3 t}^{2} - 8 t + 4 = 0 t = \frac{8 \pm \sqrt{64 - 4 \times 3 \times 4}}{2 \times 3} = \frac{8 \pm \sqrt{16}}{12} = \frac{8 \pm 4}{12} t = 1, - 2 / 3 \Rightarrow t = 1 s h = \frac{1}{2} \times {g t}^{2} = \frac{1}{2} \times 9.8 \times 1 = 4.9 m$

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