An object is thrown in vertically upward direction. The time to reach maximum height is ………..

Options

A. `(g)/(v_(0))`
B. `(v _(0) ^(2))/(g ) `
C. `(v _(0))/(g)`
D. `(v _(0))/(g ^(2))`

Answer

C

Explanation

The time to reach the maximum height: consider the vertically upward direction as positive then `v_(0)` will be positive and g will be negtive. At the maximum height, `v=0.`
` therefore` From `v =v_(0) + ` at, we get,` 0 = v _(0) - g t`
`therefore t = (v _(0))/(g)`