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`)

Options

A. `8` m
B. `4.8` m
C. `17.6` m
D. `12.8` m

Answer

A

Explanation

`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`