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A particle thrown vertically upwards has velocity 10 ms^(-1) at half of its height, then maximum height attained by it is : |
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Answer» Solution :Let u be the initial velcoity Max. height,h=`(u^(2))/(2g)` Applying `v^(2)-u^(2)=2as` Here v=10 `ms^(-1)` and `S=(h)/(2)` `100-u^(2)=-2g(h)/(2)` `100-u^(2)= -3G(h)/(2)` `100-u^(2)= -g xx(u^(2))/(2g)` or `u^(2) =200` `u=sqrt(200)` `:. h=(200)/(20)=10 m` |
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