Question

The height $h$ reached by a particle thrown up vertically as a function of times is given by $h=ut-\frac{1}{2}g{t}^2$ where $u$ and $g$ are constants. The maximum height is

• A. $\frac{u^2}{4g}$
• B. $\frac{u^2}{2g}$
• C. $\frac{u^2}{g}$
• D. $\frac{2u^2}{g}$

Answer 1

The correct option is B $\frac{u^2}{2g}$

At the maximum height the final velocity would be $0$

So equation:

$v^2-u^2=2\ast g\ast h$

v=0

$\therefore u^2=2\ast g\ast h$

$\therefore \frac{u^2}{2g}=h$