JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 16)
A small ball of mass m is thrown upward with velocity u from the ground. The ball experiences a
resistive force mkv2
where v is its speed. The maximum height attained by the ball is :
$${1 \over k}{\tan ^{ - 1}}{{k{u^2}} \over {2g}}$$
$${1 \over {2k}}{\tan ^{ - 1}}{{k{u^2}} \over g}$$
$${1 \over {2k}}\ln \left( {1 + {{k{u^2}} \over g}} \right)$$
$${1 \over k}\ln \left( {1 + {{k{u^2}} \over {2g}}} \right)$$
Explanation
_4th_September_Evening_Slot_en_16_1.png)
Fnet = ma
$$ \Rightarrow $$ -mg - mkv2 = $$mv{{dv} \over {ds}}$$
$$ \Rightarrow $$ $$ds = {{ - vdv} \over {g + k{v^2}}}$$
$$ \Rightarrow $$ $$\int\limits_{s = 0}^{{H_{\max }}} {ds} = \int\limits_{v = u}^{v = 0} {{{ - vdv} \over {g + k{v^2}}}} $$
$$ \Rightarrow $$ Hmax = $${1 \over {2k}}\ln \left( {{{g + k{u^2}} \over g}} \right)$$ = $${1 \over {2k}}\ln \left( {1 + {{k{u^2}} \over g}} \right)$$
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