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. 4 m
B. 8 m
C. 16 m
D. 10 m

Answer

B

Explanation

If we consider the given interval then work done by gravity is zero because particle is back to starting oint . Hence we can say that work is done only by air resistance .
`W_(air) = K_(2) - K_(1)`
`= 1/2 m(8)^(2) - 1/2 m(16)^(2) = 32 m - 128 m = - 96 m `
Wokr done during upward motion will be only half of this value . Let h be the maximum height attained by the aprticle .
`W_(gravity) +W_(air) = K_(2) - K_(1)`
`rArr -mgh+(-48m) = 0 - 1/2 m(16)^(2)`
` rArr " " - 10h - 48 = 0 - 128 rArr 10 h = 80 rArr h = 8 m `