A wooden block of mass `10 gm` is dropped from the top of a cliff `100 m` high. Simultaneously a bullet of same mass is fired from the foot of the cliff vertically upwards with a velocity of `100 ms^(-1)`. If the bullet after collision gets embedded in the block, the common velocity of the bullet and the block immediately after collision is `(g=10 ms^(-2))`.

Answer

(a) 4.9 m below the top and 1 s after dropping, (b) `82.4 - 4.9 = 77.5 m`

Explanation

`h = (1)/(2) g t^(2)`
`H - h = v_(1) t - (1)/(2) g t^(2)`
`H - h = v_(1) t - h`

`t = ((H)/(v_(1))) = 1 s`
`h = 4.9 m`
(b) `mv_(1) - mv_(2) = 2 m v`
`v = 40.2 m`
`h_("max") = (v^(2))/(2g) = 82.4 m`
`:.` Rise above clift `= 82.4 - 4.9 = 77.5 m`