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CBSE Class 11-science Answered

A particle is projected vertically upwards from the ground with velocity U it reaches at maximum height in time T1 after that it returns back to the ground with velocity v in time T2 if a resistance is considered then find the correct relation between T1 and t2 and correct relation between U and v?

Asked by malaysaha.80 | 01 Sep, 2019, 10:23: AM

Expert Answer

During upward motion, acceleration (dv/dt) is give by, (dv/dt) = -g - ( k v ) ...................(1)

where k is constant and v is instantaneous speed. Second term is due to air resistance and is proportional to velocity.

Eqn.(1) is written as,

begin mathsize 12px style negative fraction numerator d v over denominator g space plus space k space v end fraction space equals space d t end style

integrating both sides with proper limits, we get

begin mathsize 14px style negative integral subscript u superscript 0 fraction numerator d v over denominator g space plus space k space v end fraction space equals space integral subscript 0 superscript t subscript 1 end superscript d t 1 over k ln open parentheses fraction numerator g space plus space k space u over denominator g end fraction close parentheses space equals 1 over k ln open parentheses 1 plus fraction numerator k space u over denominator g end fraction close parentheses space space space equals space t subscript 1 space space space.............. left parenthesis 2 right parenthesis end style

During downward motion, acceleration (dv'/dt) is given by, (dv'/dt) = g - (k v' ) ..................................(3)

Eqn.(3) is written as,

begin mathsize 14px style fraction numerator d v apostrophe over denominator g minus space k space v apostrophe end fraction space equals space d t end style

integrating both sides with proper limits, we get

begin mathsize 14px style integral subscript 0 superscript v fraction numerator d v apostrophe over denominator g minus space k space v apostrophe end fraction space equals space integral subscript 0 superscript t subscript 2 end superscript d t 1 over k ln open parentheses fraction numerator g over denominator g minus k v end fraction close parentheses space space equals space minus 1 over k ln open parentheses 1 minus fraction numerator k space v over denominator g end fraction close parentheses space equals space t subscript 2 space space space................... left parenthesis 4 right parenthesis end style

From eqn.(3) and eqn.(4), we get,

begin mathsize 14px style t subscript 2 minus space t subscript 1 space equals space minus 1 over k ln open parentheses 1 space minus space fraction numerator k space v over denominator g end fraction close parentheses space minus space 1 over k space ln open parentheses 1 plus fraction numerator k space u over denominator g end fraction close parentheses space space equals space 1 over k space ln open parentheses fraction numerator 1 over denominator 1 minus begin display style fraction numerator k space v over denominator g end fraction end style end fraction cross times fraction numerator 1 over denominator 1 plus begin display style fraction numerator k space u over denominator g end fraction end style end fraction close parentheses end style

from eqn.(2), we get ,

begin mathsize 14px style u space equals space g over k open parentheses e to the power of k space t subscript 1 end exponent space minus space 1 close parentheses end style

from eqn.(4) we get,

begin mathsize 14px style v space equals space g over k left parenthesis 1 minus e to the power of negative k space t subscript 2 end exponent space right parenthesis end style

Hence,

begin mathsize 14px style v over u space equals space fraction numerator open parentheses 1 space minus space e to the power of negative k space t subscript 2 end exponent space close parentheses over denominator open parentheses e to the power of k space t subscript 1 space end exponent minus 1 close parentheses end fraction end style

Answered by Thiyagarajan K | 01 Sep, 2019, 03:46: PM

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