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a body is thrown vertically up with velocity u . it passes three points A B C in its upward journey with velocity u/2 u/3 u/4 respectively . the ratio of separation between point A & B and between B & C .I.e AB/BC is |
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Answer» answer : 20 : 7 A body is thrown vertically upward with speed u. Let point A is located at s distance from the ground initial VELOCITY = u velocity at point A ,v= u/2, using formula, v² = u² + 2as or, (u/2)² = u² + 2(-g)s or, u²/4 - u² = -2gs or, -3u²/4 = -2gs or, s = 3u²/8g .....(1) similarly, velocity at point B, v = u/3 then, (u/3)² = u² + 2(-g)s' or, -8u²/9 = -2gs' or , s' = 4u²/9g.....(2) and velocity at point C, v = u/4 then, (u/4)² = u² + 2(-g)s" or, -15u²/16 = -2gs" or, s" = 15u²/32g ....(3) so, seperation between point A and B , AB = s' - s = (4u²/9g) - (3u²/8g) = (32 - 27)u²/72g = 5u²/72g and seperation between B and C , BC = s" - s' = (15u²/32g) - (4u²/9g) = (135 - 128)u²/(288g) = 7u²/228g so, AB/BC = (5u²/72g)/(7u²/288g) = 5 × 288/72 × 7 = 20/7 |
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