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

1.	An object is thrown in vertically upward direction. The time to reach maximum height is ………..
Answer» `(g)/(v_(0))` `(v _(0) ^(2))/(g ) ` `(v _(0))/(g)` `(v _(0))/(g ^(2))` Solution :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)`

Discussion