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. $\frac{5}{7}h$

Answer 1

Answer: (B)

$\frac{5}{3}h$

Answer 2

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