InterviewSolution
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InterviewSolution
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At `t=0`, an arrow is fired vertically upwards with a speed of `100 m s^(-1)`. A second arrow is fired vertically upwads with the same speed at `t=5 s`. Then .A. The two arrows will be at the same height above the `t=20 s`,B. The two arrows will reach back their starting points at `t=20 s` and at `t=25 s`.C. The ratio of the speeds of the first and second arrow at`t= 20 s` will be `2 :1`.D. The maximum height attained by either arrow will be `1000 m`, |
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Answer» Correct Answer - A::B::C Let they meet at height `h` after time `t`. `h=100 t-(1)/(2)g t^(2) rarr` for first aarow `=100 (t-5) -(1)/(2) g (t-5)^(2) rarr` for decond arrow `rArr t =- 12.5 s` (after solving ). So (a) is correct. Time of flight of first arrow: `T=(2u)/(g) =(2xx 100)/(10) =20 s` Second arrow will reach after `5 s` of reaching first. So `b` is rossect. `{:(v_(1) =100 -10 xx 20 =- 100 m s^(-1))`,`(v_(2) =100 -10 xx 15 =-50 m s^(-1)):}` Ratio: `(v_(1))/(v_(2))=2: 1. `So (c) is correct. Maximum deight attained `H=(u^(2))/(2g)=((100)^(2))/(2xx10)=500 m`. Hence, `d` incorrect. |
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