The equivalent resistance between the point `P` and `Q` in the network given here is equal to (given `r = (3)/(2) Omega)`

Answer

`6.53 s`

Explanation

`((1)/(2) g t^(2)) + (49t - (1)/(2) t^(2)) = 98`
`rArr t = 2s u_(1) = g t(hat(j)) = -19.6hat(j)m//s`
`u_(2) = (u - g t)hat(j) = +29.4hat(j) m//s`
`v_(1) (u_(1) + u_(2))/(2) = 4.9 m//s`
`h = 4.9 xx 2 (1)/(2) xx 9.8 xx 2 xx 2 = 78.4m`
For the combined mass
`x = ut + (1)/(2) xx 9.8 t^(2), t = 4.53`
`:.` Total time of height
`= 2 + 4.53 = 6.53 sec`.