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(B) Let the total time taken to reach the ground be $t$ seconds. The initial velocity $u = 0$ and acceleration $a = g = 10 \, m/s^2$.The distance cove

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$4$

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(B) Let the total time taken to reach the ground be $t$ seconds. The initial velocity $u = 0$ and acceleration $a = g = 10 \, m/s^2$.The distance covered in the first second $(t=1)$ is given by:$x = u(1) + \frac{1}{2}g(1)^2 = 0 + \frac{1}{2} 	imes 10 	imes 1 = 5 \, m$.The distance covered in the last second is given as $7x$. Therefore,the distance in the last second is $7 	imes 5 = 35 \, m$.The formula for the distance covered in the $n^{th}$ second is $S_n = u + \frac{a}{2}(2n - 1)$.Substituting the values:$35 = 0 + \frac{10}{2}(2t - 1)$$35 = 5(2t - 1)$$7 = 2t - 1$$2t = 8$$t = 4 \, s$.Thus,the body takes $4 \, s$ to reach the ground.

$A$ body dropped from the top of a tower covers a distance $7x$ in the last second of its journey,where $x$ is the distance covered in the first second. How much time does it take to reach the ground?

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