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(D) At maximum height, the final velocity $v = 0$. Using the first equation of motion $v = u - gt$, we have $0 = u - 10 	imes 5$, which gives $u = 50

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(D) At maximum height, the final velocity $v = 0$. Using the first equation of motion $v = u - gt$, we have $0 = u - 10 	imes 5$, which gives $u = 50 \,m/s$.Distance travelled in the $n^{th}$ second is given by $s_n = u + \frac{a}{2}(2n - 1)$.For the $2^{nd}$ second $(n=2)$: $s_2 = 50 - \frac{10}{2}(2 	imes 2 - 1) = 50 - 5(3) = 35 \,m$.For the $7^{th}$ second $(n=7)$: $s_7 = 50 - \frac{10}{2}(2 	imes 7 - 1) = 50 - 5(13) = 50 - 65 = -15 \,m$. The magnitude of distance is $15 \,m$.The ratio of distance in the $2^{nd}$ second to the $7^{th}$ second is $\frac{35}{15} = \frac{7}{3}$.

$A$ ball is thrown vertically upwards with an initial velocity $u$ and reaches its maximum height in $5 \,s$. The ratio of the distance travelled by the ball in the $2^{nd}$ and $7^{th}$ second is (assume $g=10 \,m/s^2$):

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