1.

An object is thrown in vertical upward direction with velocity 'u' then the maximum height attained by the object would be....

Answer»

`(u)/(G)`
`(u ^(2))/(2g)`
`(u ^(2))/(g)`
`(u)/(2g)`

Solution :`2AS =V ^(2) -u^(2)`
At maximum height, the final VELOCITY of the object v=0 and for motion in upward direction `a =-g `
`therefore 2 (-g) H _(max) = 0 -u^(2) (because s = h _(max))`
`therefore h _(max) = (u ^(2))/(2g)`


Discussion

No Comment Found

Related InterviewSolutions