A particle is dropped under gravity from rest | vedclass

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Question

(B) Let the total time taken to fall from height $h$ be $n$ seconds.Using the equation of motion for distance covered from rest: $h = \frac{1}{2}gn^2$

Vedclass · Accepted Answer

$122.5$

Vedclass · Answer

(B) Let the total time taken to fall from height $h$ be $n$ seconds.Using the equation of motion for distance covered from rest: $h = \frac{1}{2}gn^2$ ... $(i)$Distance covered in the last second ($n^{th}$ second) is given by $S_n = \frac{g}{2}(2n - 1)$.According to the problem,$S_n = \frac{9h}{25}$.Substituting the value of $h$ from $(i)$ into this expression:$\frac{9}{25} \left( \frac{1}{2}gn^2 ight) = \frac{g}{2}(2n - 1)$$\frac{9n^2}{25} = 2n - 1$$9n^2 = 50n - 25$$9n^2 - 50n + 25 = 0$Solving the quadratic equation: $9n^2 - 45n - 5n + 25 = 0$$9n(n - 5) - 5(n - 5) = 0$$(9n - 5)(n - 5) = 0$Since $n$ must be greater than $1$ second for the last second to exist,we take $n = 5 \ s$.Now,substitute $n = 5$ into equation $(i)$:$h = \frac{1}{2} 	imes 9.8 	imes (5)^2$$h = 4.9 	imes 25 = 122.5 \ m$.

$A$ particle is dropped under gravity from rest from a height $h$ $(g = 9.8 \ m/s^2)$. If it travels a distance of $9h/25$ in the last second of its motion,the height $h$ is ......... $m$.

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