A ball is projected upwards from a height h above the surface of the earth with velocity v. The time at which the ball strikes the ground is

Options

A. `v/g+(2hg)/sqrt2`
B. `v/g[1-sqrt(1+"2h"/g)]`
C. `v/g[1+sqrt(1+"2gh"/v^2)]`
D. `v/g [ 1+sqrt(v^2 + "2g"/v^2)]`

Answer

C

Explanation

Since direction of v is opposite to the direction of g and h so from equation of motion
`h=-vt + 1/2 "gt"^2`
`rArr "gt"^2 - 2vt - 2h =0`
`rArr t=(2v pm sqrt(4v^2 + 8gh))/"2g"`
`rArr t=v/g [1+sqrt(1+"2gh"/v^2)]`