Question

A particle is projected vertically upwards and it attains maximum height $H$. If the ratio of times to attain height $h(h<H)$ is $1/3$, then $h$ equals

• A. 2/3 H
• B. 3H
• C. 5/9. H
• D. 3/. H

Answer 1

The correct option is C 5/9. H

Total time taken to reach height H$=t=\sqrt{\frac{2H}{g}}$

Initial velocity$=u=\sqrt{2gH}$

Time to reach height h$=t_1=t/3=\frac{1}{3}\sqrt{\frac{2H}{g}}$

Now by second equation of motion

$\Rightarrow h=u{t}_1+0.5\times (-g)\left(t_1^2\right)$

$\Rightarrow h=\sqrt{2gH}\times$$\frac{1}{3}\times \sqrt{\frac{2H}{g}}$$-0.5\times g\times {\left(\frac{1}{3}\sqrt{\frac{2H}{g}}\right)}^2$

$\Rightarrow h=\frac{2H}{3}-\frac{H}{9}=\frac{5H}{9}$

$\Rightarrow \frac{h}{H}=\frac{5}{9}$

Hence none of the options are correct.