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(B) The distance travelled in the $n^{th}$ second is given by $S_n = u + \frac{g}{2}(2n - 1)$.Since the body is moving upwards,the acceleration due to

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$49$

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(B) The distance travelled in the $n^{th}$ second is given by $S_n = u + \frac{g}{2}(2n - 1)$.Since the body is moving upwards,the acceleration due to gravity is $-g$. Thus,$S_n = u - \frac{g}{2}(2n - 1)$.For the $5^{th}$ second $(n=5)$: $S_5 = u - \frac{g}{2}(2(5) - 1) = u - 4.5g$.For the $6^{th}$ second $(n=6)$: $S_6 = u - \frac{g}{2}(2(6) - 1) = u - 5.5g$.However,the distance travelled is the magnitude of displacement. If the body reaches its highest point between the $5^{th}$ and $6^{th}$ second,the displacement in the $5^{th}$ second is positive and in the $6^{th}$ second is negative.Given $|S_5| = |S_6|$,we have $u - 4.5g = -(u - 5.5g)$.$u - 4.5g = -u + 5.5g$.$2u = 10g$.$u = 5g = 5 	imes 9.8 = 49\, m/s$.

$A$ body is thrown vertically upwards with velocity $u$. The distance travelled by it in the fifth and the sixth seconds are equal. The velocity $u$ is given by.........$m/s$ $(g = 9.8\, m/s^{2})$

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