InterviewSolution
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A particle is projected at an angle of `45^(@)` with a velocity of `9.8 ms^(-1)`. The horizontal range will be (Take, `g = 9.8 ms^(-2))`A. 9.8mB. 4.9mC. 9.8/ `sqrt(2)m`D. `9.8sqrt(2)m` |
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Answer» Correct Answer - a `R=(u^(2)sin2theta)/(g)` |
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