Question

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

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

Answer 1

Answer: (B)

10 m

Answer 2

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