In the figure,the mass of a ball is frac{9}{5 | vedclass

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(A) Let $a_{1}$ and $a_{2}$ be the accelerations of the ball (upward) and the rod (downward),respectively.From the constraint relation,$2 a_{1} = a_{2

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$1.4$

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(A) Let $a_{1}$ and $a_{2}$ be the accelerations of the ball (upward) and the rod (downward),respectively.From the constraint relation,$2 a_{1} = a_{2} \dots (i)$For the ball,the equation of motion is: $2T - \frac{9}{5}mg = \frac{9}{5}ma_{1} \dots (ii)$For the rod,the equation of motion is: $mg - T = ma_{2} \dots (iii)$Substituting $T = mg - ma_{2}$ from $(iii)$ into $(ii)$:$2(mg - ma_{2}) - \frac{9}{5}mg = \frac{9}{5}ma_{1}$$2mg - 2ma_{2} - 1.8mg = 1.8ma_{1}$$0.2g - 2a_{2} = 1.8a_{1}$Using $a_{2} = 2a_{1}$ from $(i)$:$0.2g - 2(2a_{1}) = 1.8a_{1}$$0.2g = 5.8a_{1} \implies a_{1} = \frac{0.2g}{5.8} = \frac{g}{29} \, m/s^2$ (upward)Then $a_{2} = 2a_{1} = \frac{2g}{29} \, m/s^2$ (downward)The relative acceleration of the ball with respect to the rod is $a_{rel} = a_{1} + a_{2} = \frac{g}{29} + \frac{2g}{29} = \frac{3g}{29}$.Using the equation of motion $s = ut + \frac{1}{2}a_{rel}t^2$ with $u=0$ and $s=1 \, m$:$1 = 0 + \frac{1}{2} \left(\frac{3 	imes 10}{29}ight) t^2$$1 = \frac{15}{29} t^2 \implies t^2 = \frac{29}{15} \approx 1.933$$t = \sqrt{1.933} \approx 1.39 \, s \approx 1.4 \, s$.

In the figure,the mass of a ball is $\frac{9}{5}$ times the mass of the rod. The length of the rod is $1 \, m$. The initial level of the ball is the same as the lower end of the rod. Find the time taken by the ball to reach the upper end of the rod. (in $s$)

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