A particle projected vertically upwards attains a maximum height H. If the ratio of the time to attain a height h (h<H) to the time to attain height H is 1/3 then

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9h = 5H

3h = H

3h = 4H

4h = H

answer is A.

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Detailed Solution

$\begin{array}{l} 0^{2} - u^{2} = 2 (- g) H \\ \Rightarrow u = \sqrt{2 gH} . . . . (i) \\ T i m e t a k e n t o r e a c h H i s t_{2} a n d T i m e t a k e n t o r e a c h h i s t_{1} \\ t_{1} = \frac{t_{2}}{3} = \frac{1}{3} \sqrt{\frac{2 H}{g}} (∵ \frac{t_{1}}{t_{2}} = \frac{1}{3}) \\ \Rightarrow h = {ut}_{1} - \frac{1}{2} {gt}_{1}^{2} \\ \Rightarrow h = (\sqrt{2 gH}) \frac{1}{3} \sqrt{\frac{2 H}{g}} - \frac{1}{2} g {(\frac{1}{3} \sqrt{\frac{2 H}{g}})}^{2} \\ h = \frac{2 H}{3} - \frac{H}{9} \\ \frac{h}{H} = \frac{5}{9} \end{array}$