A body is dropped from a height h. It covers | vedclass

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(B) Let the total time taken to reach the ground be $n$ seconds.The total distance covered is $h = \frac{1}{2}gn^2$  ---$(1)$The distance covered in t

Vedclass · Accepted Answer

$122.5$

Vedclass · Answer

(B) Let the total time taken to reach the ground be $n$ seconds.The total distance covered is $h = \frac{1}{2}gn^2$  ---$(1)$The distance covered in the last second is given by $S_n = \frac{g}{2}(2n - 1)$.According to the problem,$S_n = \frac{9h}{25}$.Substituting $h$ from equation $(1)$ into the expression for $S_n$:$\frac{g}{2}(2n - 1) = \frac{9}{25} 	imes (\frac{1}{2}gn^2)$$(2n - 1) = \frac{9n^2}{25}$$50n - 25 = 9n^2$$9n^2 - 50n + 25 = 0$Solving the quadratic equation using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$n = \frac{50 \pm \sqrt{2500 - 900}}{18} = \frac{50 \pm \sqrt{1600}}{18} = \frac{50 \pm 40}{18}$Taking the positive value,$n = \frac{90}{18} = 5 \, s$.Now,substitute $n = 5$ into equation $(1)$ with $g = 9.8 \, m/s^2$:$h = \frac{1}{2} 	imes 9.8 	imes (5)^2 = 4.9 	imes 25 = 122.5 \, m$.

$A$ body is dropped from a height $h$. It covers a distance of $9h/25$ in the last second of its motion. What is the height $h$ in $m$?

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