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An object is projected with a velocity of `10 m//s` at an angle `45^(@)` with horizontal. The equation of trajectory followed by the projectile is `y = a x - beta x^(2)`, the ratio `alpha//beta` isA. `5`B. `10`C. `15`D. `20` |
Answer» Correct Answer - B `y = alpha x - beta x^(2) = x tan theta - (g x^(2))/(2u^(2) cos^(2) theta)` `alpha = tan theta, beta = (g)/(2u^(2) cos^(2) theta)` `(alpha)/(beta) = (tan theta)/(g//2u^(2) cos^(2) theta)` `= (2u^(2) cos theta sin theta)/(g) = (2u^(2) cos^(2) theta tan theta)/(g)` `= ((10)^(2))/(10)sin 90^(@) = 10` |
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