Question

A particle is thrown vertically upwards. Its velocity at half of the height is $10m{s}^{-1}$ . Then the maximum height attained by it is
(Take $g=10m{s}^{-2}$ )

• A. 16 m
• B. 10 m
• C. 8 m
• D. 18 m

Answer 1

Answer: (B)

At maximum height vertical component of final velocity is zero.
It is given velocity at half the height is $10m/s$.
From equation of motion, we have
$v^2=u^2-2gs$
where $v$ is final velocity,
$g$ is acceleration due to gravity and $s$ is displacement.
At maximum height $v=0$
$\therefore u^2=2gs$
$\Rightarrow s=\frac{u^2}{2g}$
At half the height,
$\Rightarrow s^{{}^{\prime }}=\frac{s}{2}=\frac{1}{2}\left(\frac{u^2}{2g}\right)$
Now $100-u^2=2\times (-g)\times \frac{u^2}{4g}$
$\Rightarrow u=\sqrt{200}m/s$
Maximum height attained is
$=\frac{200}{(2\times 10)}=10m$