Question

From an elevated point $P$, a stone is projected vertically upwards. When the stone reaches a distance $h$ below $P$, its velocity is double of its velocity at a height $h$ above $P$. The greatest height attained by the stone from the point of projection $P$ is :

• A. $\frac{3}{5}h$
• B. $\frac{5}{3}h$
• C. $\frac{7}{5}h$
• D. $\frac{5}{7}h$
• E. $\frac{2}{3}h$

Answer 1

The correct option is B $\frac{5}{3}h$

From equations of motions, we know
$v^2=u^2-2gh$ ...(i)
Here, $v=2v$, so
$(2v{)}^2=u^2+2gh$ ...(ii)
Now, on adding Eqs. (i) and (ii), we get
$5v^2=2u^2$
$u^2=\frac{5}{2}{v}^2$ ...(iii)
On subtracting Eq. (i) in Eq. (ii), we get
$3v^2=4gh$
$v^2=\frac{4}{3}gh$ ...(iv)
And we know that the
$H=\frac{u^2}{2g}$
So, here $H=\frac{\frac{5}{2}{v}^2}{2g}=\frac{\frac{5}{2}\left(\begin{array}{c}\frac{4}{3}gh\end{array}\right)}{2g}$
$H=\frac{5}{3}h$