Question

A ball is thrown up with a certain velocity so that it reaches a height h. Find the ratio of the times in which it is a h/3.

• A. $\frac{\sqrt{2} - 1}{\sqrt{2} + 1}$
• B. $\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$
• C. $\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
• D. $\frac{1}{3}$

Answer 1

Answer: (B)

$\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

Answer 2

u² = 2gh; h/3 = $\sqrt{2gh}$ t – 1/2 gt² or gt² – 2

$\sqrt{2gh}$t + 2h/3 = 0

t = $\frac{2\sqrt{2gh} \pm \sqrt{8gh - (8gh)/3}}{2g}$

or $\frac{t_{1}}{t_{2}} = \frac{2\sqrt{2gh} - 2\sqrt{2gh/3}\left( \sqrt{3 - 1} \right)}{2\sqrt{2gh} + 2\sqrt{2gh/3}\left( \sqrt{3 - 1} \right)}$

= $\frac{\sqrt{3} - \left( \sqrt{3 - 1} \right)}{\sqrt{3} + \left( \sqrt{3 - 1} \right)}$ = $\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$