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The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmospehre) are given by `y=(8t-5t^(2))` metre and x=6t metre, where t is in seconds. The velocity of projection is:A. 8m/sB. 6m/sC. 10m/sD. not obtianed from the data |
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Answer» Correct Answer - c `y=(u_(x))/(3)-(5x^(2))/(36)` Now, `y=x tan theta-(gx^(2))/(2u^(2)cos^(2)theta)` `u=10m//s` |
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