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A body is projected at an angle `theta` so that its range is maximum.If `T` is the time of flight then the value of maximum range is (acceleration due to gravity =`g`)A. `(g^(2)T)/2`B. `(gT)/2`C. `(gT^(2))/2`D. `(g^(2)T^(2))/2` |
Answer» Correct Answer - C for maximum range `theta=45^(@)` `R_(max)=u^(2)/g` when`theta=45^(@),T=(2usintheta)/g` |
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