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In horizontal level ground to projectile if at any instant velocity becomes perpendicular to initial velocity then what can you say about projection angle with horizontal.A. `theta=45^(@)`B. `thetage45^(@)`C. `thetale45^(@)`D. for any value of `theta` it is possible |
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Answer» Correct Answer - b Velocity at any time `vecv=vecu+vec(g)t` `implies vecv=u cos theta hati+(u sin theta-gt)hatj` Let the any time this velocity becomes perpendicular to initial velocity. The `vecv.vecu=0` Solve to get `t=u/(g sin theta)` Now t should be less then/equal to time of flight. So, `tgeT`. `u/(g sin theta)le(2u sin theta)/g implies sin^(2) thetage1/2` `implies sin thetage1/(sqrt(2)) implies thetage45^(@)` |
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