First slide

Conservation of mechanical energy

Question

A particle is projected vertically upwards with a speed of 16 ms^-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 ms^-2)

Moderate

A

8 m
B

4.8 m
C

17.6 m
D

12.8 m

Solution

From work-energy theorem, for upward motion
$\frac{1}{2} m (16)^{2} = mgh + W$ (work done by air resistance)
for downward motion,
$\frac{1}{2} m (8)^{2} = mgh - W \Rightarrow \frac{1}{2} [(16)^{2} + (8)^{2}] = 2 gh$
or h = 8m

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