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A particle is projected vertically upwards with a speed of `16ms^-1`. After some time, when it again passes through the point of projection, its speed is found to be `8ms^-1`. It is known that the work done by air resistance is same during upward and downward motion. Then the maximum height attained by the particle is (take `g=10ms^-2`)A. `8` mB. `4.8` mC. `17.6` mD. `12.8` m |
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Answer» Correct Answer - A `W_(g)+W_(air)=DeltaKE` during upward `-mgh-((W_(air))/(2))=0-(1)/(2)m(16)^(2)` during downward `+mgh-((W_(air))/(2))=(1)/(2)m(8)^(2)` subtracting `2mgh=(1)/(2)m(8^(2)+16^(2))` so `h=(64+256)/(40)=8m` |
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