If the maximum possible exhaust velocity of a rocket be `2km//s` calculate the ratio `m_(0)//m` for it to acquire a veocity of `11.2km//s` after starting from rest .

Answer

`(m_(0))/(m) =270.4` .

Explanation

Here, `u = 2 km//s, upsilon = 11.2 km//s, ((m_(0))/(m)) =` ?
As rocket starts from rest therefore,
`upsilon = ulog_(e) ((m_(0))/(m)) = u xx 2.303log_(10)((m_(0))/(m))`
`log_(10)((m_(0))/(m)) = (upsilon)/(2.303 u) = (11.2)/(2.303 xx2) = 2432`
`m_(0)/(m) = "antilog" 2.432 = 270.4` .