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A particle is projected horizontally with a speed of `(40)/(sqrt3)m//s`, from some height `t = 0`. At what time its velocity makes `60^(@)` angle with initial velocity. `(g=10m//s^(2))` A. 4sB. 2sC. 5sD. 3s |
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Answer» Correct Answer - A `tan alpha=(v_(y))/(v_(x)),sqrt3=("gt")/(((40)/(sqrt3))), sqrt3=(g t sqrt3)/(40 t=4s` |
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