MCQ
A particle is projected vertically upwards with a speed of $16\  m/s$ , after some time , when it again passes through the point of projection, its speed is found to be $8\  m/s$ . 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 ...................... $\mathrm{m}$ ( $g$ = $10\  m/s^2$ )
  • $8$
  • B
    $4.8$
  • C
    $17.6$
  • D
    $12.8$

Answer

Correct option: A.
$8$
a
$\mathrm{W}_{\mathrm{s}}+\mathrm{W}_{\mathrm{air}}=\Delta \mathrm{KE}$

during upward

$-m g h-\left(\frac{W_{\text {air }}}{2}\right)=0-\frac{1}{2} m(16)^{2}$

during downward

$+m g h-\left(\frac{W_{\text {air }}}{2}\right)=\frac{1}{2} m(8)^{2}$

subtracting

$2 \mathrm{mgh}=\frac{1}{2} \mathrm{m}\left(8^{2}+16^{2}\right)$

$\mathrm{h}=\frac{64+256}{4 \mathrm{g}}=8$

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