Question

Derive an expression for man height attained by an object projected vertically upwards with a velocity 'u'

Answer 1

Solution

REF.Image

At max height (Let H):

$v=0m/s$. so,

Applying third $e{q}^n$ of motion :

$v^2-u^2=2as$

Here, $a=-g$ and $S=H$. So,

$(0{)}^2-(u{)}^2=2(-g)H$

$\Rightarrow \frac{u^2}{2g}=H$