When a ball is thrown up vertically with velocity `v_(0)` it reaches a maximum height of h. when the ball reaches to triple t he maximum height. The velocity with which the ball was thrown is :

Options

A. `3v_(0)`
B. `9v_(0)`
C. `sqrt(3)v_(0)`
D. `2v_(0)`

Answer

C

Explanation

`v^(2) - u^(2) = 2as`
Here ` u = v_(0), s = h, v = 0 , a = - g`
`0^(2) - v_(0)^(2) - 2 (-g)h`
`:. v_(0)^(2) = 2gh " "i.e., v_(0)^(2) prop h`
i,e.,` v_(0)^(2) prop 3h`
`(v'_(0)^(2))/(v'_(0)^(2))= (3h)/(h) = 3 `
`:. v'_(0)^(2)= 3 v_(0)^(2) `
`:. v'_(0) = sqrt(3) v_(0)`