Question

A body is thrown vertically up with certain velocity. If $h$ is the maximum height reached by it, its position when its velocity reduces to $(1/3{)}^{rd}$ of its velocity of projection is at

• A. 8h/9 from the ground
• B. 8h/9 below the top-most point
• C. 4h/9 from the ground
• D. h/3 below the top-most point

Answer 1

The correct option is A 8h/9 from the ground

Let initial velocity is $u$,
$h=\frac{u_1^2}{2g}$

$v^2=u^2-2gx$

$(\frac{u_1}{3}{)}^2=u_1^2-2gx$

$2gx=\frac{8}{9}{u}_1^2$

$x=\frac{8}{9}\left(\frac{u_1^2}{2g}\right)=\frac{8}{9}h$