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(B) Given,time of ascent $t = 6 \; s$. Since $t = u/g$,we have $u = 6 	imes 10 = 60 \; m/s$ (taking $g = 10 \; m/s^2$).Distance travelled in the $n^{

Vedclass · Accepted Answer

$11:1$

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(B) Given,time of ascent $t = 6 \; s$. Since $t = u/g$,we have $u = 6 	imes 10 = 60 \; m/s$ (taking $g = 10 \; m/s^2$).Distance travelled in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.For the first second $(n = 1)$: $S_1 = 60 + \frac{-10}{2}(2(1) - 1) = 60 - 5 = 55 \; m$.For the seventh second $(n = 7)$: $S_7 = 60 + \frac{-10}{2}(2(7) - 1) = 60 - 5(13) = 60 - 65 = -5 \; m$.The magnitude of distance is $5 \; m$.The ratio of the distances is $55:5 = 11:1$.

$A$ body thrown vertically upwards with an initial velocity $u$ reaches maximum height in $6$ seconds. The ratio of the distances travelled by the body in the first second and the seventh second is

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