Question

A 0.5 kg ball is thrown up with an initial speed 14 m/s and reaches maximum height of 8.0 m how much energy dissipated by air drag acting on the ball during the ascent

Answer 1

If there is no air drag then maximum height,                

H=u^2/2g=14×14/2×9.8=10m

But due to air drag ball reaches up to height 8m only.

So, loss in energy                 

=mg(10−8)

=0.5×9.8×2=9.8J

Answer 2

Hope it helps u......