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

Vedclass · Accepted Answer

$49$

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(C) The distance traveled in the $n^{th}$ second is given by $S_n = u + \frac{g}{2}(2n - 1)$.Since the particle is moving upwards,the acceleration due to gravity is $-g$.Thus,$S_n = u - \frac{g}{2}(2n - 1)$.Given that the distance traveled in the $5^{th}$ second is equal to the distance traveled in the $6^{th}$ second,we have:$u - \frac{g}{2}(2(5) - 1) = |u - \frac{g}{2}(2(6) - 1)|$.Since the particle reaches its maximum height and starts coming down,the displacement in the $6^{th}$ second must be negative of the displacement in the $5^{th}$ second if the distances are equal.Alternatively,the particle reaches the peak between $t=5$ and $t=6$. The peak occurs at $t = u/g$.For the distances to be equal in the $5^{th}$ and $6^{th}$ seconds,the peak must occur at $t = 5.5$ seconds.$5.5 = u / 9.8$$u = 5.5 	imes 9.8 = 53.9 \, m/s$.Wait,re-evaluating: If the particle travels equal distance in $5^{th}$ and $6^{th}$ second,it means it crosses the peak. The distance in $n^{th}$ second is $S_n = u + a(n - 0.5)$.For $n=5$,$d_5 = u - 4.5g$. For $n=6$,$d_6 = -(u - 5.5g)$.$u - 4.5g = -u + 5.5g$$2u = 10g$$u = 5g = 5 	imes 9.8 = 49 \, m/s$.

For a particle projected vertically upwards under gravity,it travels equal distance during the $5^{th}$ and $6^{th}$ second of its motion. Find its projection speed in $m/s$ $(g = 9.8\,m/s^2)$.

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