A stone with weight w is thrown vertically up | vedclass

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(B) According to the work-energy theorem,the net work done on the object is equal to the change in its kinetic energy: $W_{	ext{net}} = \Delta K$.Dur

Vedclass · Accepted Answer

$h=\frac{v_0^2}{2 g\left(1+\frac{f}{w}\right)}$

Vedclass · Answer

(B) According to the work-energy theorem,the net work done on the object is equal to the change in its kinetic energy: $W_{	ext{net}} = \Delta K$.During the upward motion,both the gravitational force $(w = mg)$ and the air drag force $(f)$ act downwards,opposing the motion.The work done by gravity is $W_g = -wh = -mgh$.The work done by air drag is $W_f = -fh$.The change in kinetic energy is $\Delta K = K_f - K_i = 0 - \frac{1}{2}mv_0^2$.Applying the theorem: $W_g + W_f = \Delta K$.$-mgh - fh = -\frac{1}{2}mv_0^2$.Dividing by $m$: $-gh - \frac{f}{m}h = -\frac{1}{2}v_0^2$.Since $w = mg$,we have $m = \frac{w}{g}$. Substituting this into the drag term: $-gh - \frac{fg}{w}h = -\frac{1}{2}v_0^2$.$h(g + \frac{fg}{w}) = \frac{1}{2}v_0^2$.$h g (1 + \frac{f}{w}) = \frac{1}{2}v_0^2$.$h = \frac{v_0^2}{2g(1 + \frac{f}{w})}$.

$A$ stone with weight $w$ is thrown vertically upward into the air from ground level with initial speed $v_0$. If a constant force $f$ due to air drag acts on the stone throughout its flight,the maximum height attained by the stone is

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