A body of mass 2 kg is thrown up vertically with K.E. of 490 joule. If the acceleration due to gravity is 9.8 $m / s^{2}$ , then the height at which the K.E. of the body becomes half its original value is given by

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

The correct option is D 9.8 JouleIf there is no air drag then maximum height, H=u22g=14×142×9.8=10mBut 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

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

The correct option is D $9.8$ Joule
If there is no air drag then maximum height,
$H = \frac{u^{2}}{2 g} = 14 \times \frac{14}{2} \times 9.8 = 10 m$
But due to air drag ball reaches up to height 8m only.
So, loss in energy
$= m g (10 - 8)$
$= 0.5 \times 9.8 \times 2 = 9.8 J$