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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 isA. `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)]` |
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Answer» Correct Answer - C 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)]` |
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