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  • A ball is released from a height . If  <img alt="\mathrm{t_{1} \: and \: t_{2}}" src="https:

A ball is released from a height \mathrm{h}. If  \mathrm{t_{1} \: and \: t_{2}}  be the time required to complete the first half and second half of the distance respectively. Then, choose the correct relation between  \mathrm{ t_{1} \: and \: t_{2}}.

 

Option: 1

\mathrm{{t_{1}}=(\sqrt{2}) t_{2}}


Option: 2

\mathrm{t_{1}=(\sqrt{2}-1) t_{2} }


Option: 3

\mathrm{t_{2}=(\sqrt{2}+1) t_{1}}
 


Option: 4

\mathrm{t_{2}=(\sqrt{2}-1) t_{1}}


Answers (1)

best_answer

From A to B

\mathrm{S=u t+\frac{1}{2} a t^2 }

\mathrm{-\frac{h}{2}=0+\frac{1}{2}(-g) t_1^2 }

\mathrm{t_1=\sqrt{\frac{h}{g}} \rightarrow(1) }

From A to C

\mathrm{s =u t+\frac{1}{2} a t^2 }

\mathrm{-h=0+\frac{1}{2}(-g)\left(t_1+t_2\right)^2 }

\mathrm{t_1+t_2 =\sqrt{\frac{2 h}{g}} \rightarrow(2) }

From eq (1) and (2)

\mathrm{t_2 =\sqrt{\frac{2 h}{g}}-\sqrt{\frac{h}{g}} \rightarrow(3) }

\mathrm{\frac{t_1}{t_2} =\frac{1}{\sqrt{2}-1} }

\mathrm{t_2 =(\sqrt{2}-1) t_1}

Posted by

shivangi.bhatnagar

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