The distance covered by a stone dropped from | vedclass

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(C) Let the total time of motion be $t$ seconds. The total height of the building $H$ is given by $H = \frac{1}{2}gt^2$.The distance covered in the la

Vedclass · Accepted Answer

$122.5$

Vedclass · Answer

(C) Let the total time of motion be $t$ seconds. The total height of the building $H$ is given by $H = \frac{1}{2}gt^2$.The distance covered in the last second is the difference between the total distance and the distance covered in $(t-1)$ seconds.Distance in last second $= H - \frac{1}{2}g(t-1)^2$.According to the problem,$H - \frac{1}{2}g(t-1)^2 = 0.36H$.Substituting $H = \frac{1}{2}gt^2$,we get $\frac{1}{2}gt^2 - \frac{1}{2}g(t-1)^2 = 0.36 	imes (\frac{1}{2}gt^2)$.Dividing by $\frac{1}{2}g$,we get $t^2 - (t-1)^2 = 0.36t^2$.$t^2 - (t^2 - 2t + 1) = 0.36t^2$.$2t - 1 = 0.36t^2$.$0.36t^2 - 2t + 1 = 0$.Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $t = \frac{2 \pm \sqrt{4 - 4(0.36)(1)}}{2(0.36)} = \frac{2 \pm \sqrt{4 - 1.44}}{0.72} = \frac{2 \pm \sqrt{2.56}}{0.72} = \frac{2 \pm 1.6}{0.72}$.Taking the positive value,$t = \frac{3.6}{0.72} = 5 \ s$.Now,$H = \frac{1}{2} 	imes 9.8 	imes (5)^2 = 4.9 	imes 25 = 122.5 \ m$.

The distance covered by a stone dropped from the top of a building in the last second of its motion is $0.36$ times the total distance travelled by it. The height of the building is (acceleration due to gravity $g = 9.8 \ m/s^2$). (in $m$)

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