A particle is projected vertically upwards with a speed of 16 m s^-1. After some time, when it again passes through the point of projection, its speed is found to be 8 ms^-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 = 10 m s^-2)

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12.8 m

17.6 m

8 m

4.8 m

answer is A.

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Detailed Solution

From work-energy theorem, for upward motion
$\frac{1}{2} m (16)^{2} = mgh + W (work done by air resistance)$

for downward motion,

$\begin{array}{l} \frac{1}{2} m (8)^{2} = mgh - W \Rightarrow \frac{1}{2} [(16)^{2} + (8)^{2}] = 2 gh \\ or h = 8 m \end{array}$

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