A mass $m$ is thrown vertically upward into air with initial speed $u$ . A constant force $F$ due to air resistance acts on the mass during it's travel. Taking into account the work done against air drag, the maximum distance covered by the mass to reach the top is (Given acceleration due to gravity $= g$ )

\frac{u^{2}}{2 g}

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\frac{u^{2}}{2 g + 2 F / m}

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\frac{u^{2}}{2 g + F / m}

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\frac{u^{2}}{g + F / m}

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Solution

The correct option is A $\frac{u^{2}}{2 g + 2 F / m}$
Loss in K.E. $=$ Gain in P.E. + Work done against air drag
i.e. $\frac{1}{2} m u^{2} = m g h + F h = h (m g + F)$
or $h = \frac{u^{2}}{2 (g + F / m)}$
$= \frac{u^{2}}{2 g + 2 F / m}$

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If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$ , then $h$ is equal to $\frac{u^{2}}{2 g}$ . [ $g$ is the acceleration due to gravity]

Q. Assertion :Maximum possible height attained by a projectile is

\frac{u^{2}}{2 g}

, where u is initial velocity Reason: To attain maximum height, a particle is thrown vertically upwards at

θ = 90^{o}

to the ground.

H = \frac{u^{2}}{2 g} s i n^{2} θ \Rightarrow H_{m a x} = \frac{u^{2}}{2 g}

Q. Question 2
A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a)

\frac{u}{2}

(b)

\frac{u^{2}}{2 g}

(c)

\frac{u^{2}}{g}

(d)

\frac{u}{2 g}

If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$ , then $h$ is equal to $\frac{u^{2}}{2 g}$ :

Q. Question 2
A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a)

\frac{u}{2}

(b)

\frac{u^{2}}{2 g}

(c)

\frac{u^{2}}{g}

(d)

\frac{u}{2 g}

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A mass m is thrown vertically upward into air with initial speed u. A constant force F due to air resistance acts on the mass during it's travel. Taking into account the work done against air drag, the maximum distance covered by the mass to reach the top is (Given acceleration due to gravity =g)

The correct option is A u22g+2F/mLoss in K.E. = Gain in P.E. + Work done against air dragi.e. 12mu2=mgh+Fh=h(mg+F)or h=u22(g+F/m)=u22g+2F/m

The correct option is A $\frac{u^{2}}{2 g + 2 F / m}$
Loss in K.E. $=$ Gain in P.E. + Work done against air drag
i.e. $\frac{1}{2} m u^{2} = m g h + F h = h (m g + F)$
or $h = \frac{u^{2}}{2 (g + F / m)}$
$= \frac{u^{2}}{2 g + 2 F / m}$