moderate
QJEEM20Ph77b7: JEE Main 2020 Physics - Class 11 JEEM20Ph77b7 Solution: A small ball of mass m is thrown upward with velocity u from…
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 :
Multiple Choice Options
A
1 k tan − 1 k u 2 2 g {1 \over k}{\tan ^{ - 1}}{{k{u^2}} \over {2g}} k 1 tan − 1 2 g k u 2
B
1 2 k tan − 1 k u 2 g {1 \over {2k}}{\tan ^{ - 1}}{{k{u^2}} \over g} 2 k 1 tan − 1 g k u 2
C
1 2 k ln ( 1 + k u 2 g ) {1 \over {2k}}\ln \left( {1 + {{k{u^2}} \over g}} \right) 2 k 1 ln ( 1 + g k u 2 )
D
1 k ln ( 1 + k u 2 2 g ) {1 \over k}\ln \left( {1 + {{k{u^2}} \over {2g}}} \right) k 1 ln ( 1 + 2 g k u 2 )
Step-by-Step Verified Solution
F
net = ma
⇒ \Rightarrow ⇒ -mg - mkv
2 =
m v d v d s mv{{dv} \over {ds}} m v d s d v ⇒ \Rightarrow ⇒ d s = − v d v g + k v 2 ds = {{ - vdv} \over {g + k{v^2}}} d s = g + k v 2 − v d v ⇒ \Rightarrow ⇒ ∫ s = 0 H max d s = ∫ v = u v = 0 − v d v g + k v 2 \int\limits_{s = 0}^{{H_{\max }}} {ds} = \int\limits_{v = u}^{v = 0} {{{ - vdv} \over {g + k{v^2}}}} s = 0 ∫ H m a x d s = v = u ∫ v = 0 g + k v 2 − v d v ⇒ \Rightarrow ⇒ H
max =
1 2 k ln ( g + k u 2 g ) {1 \over {2k}}\ln \left( {{{g + k{u^2}} \over g}} \right) 2 k 1 ln ( g g + k u 2 ) =
1 2 k ln ( 1 + k u 2 g ) {1 \over {2k}}\ln \left( {1 + {{k{u^2}} \over g}} \right) 2 k 1 ln ( 1 + g k u 2 )
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