Question

A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is $10m{s}^{-1}$ , then the maximum height attained by the stone is $(g=10m{s}^{-2})$

• A. 5 m
• B. 150 m
• C. 20 m
• D. 10 m

Answer 1

Answer: (D)

Let $u$ be the initial velocity and $h$ be the maximum height attained by the stone.
$v_1^2=u^2-2g{h}_1$
$\therefore (10{)}^2=u^2-2\times 10\frac{h}{2}$
$\left(\because h_1=\frac{h}{2},v_1=10m{s}^{-1}\right)$
or $100=u^2-10h$ ...(i)
Again at height $h$
$v_2^2=u^2-2gh$
$0=u^2-2\times 10\times h\left(\because v_2=0\right)$
$0=u^2-20h$ ...(ii)
Subtracting Eq. (ii) from Eq. (i), we get
or $100=10h$
$h=\frac{100}{10}=10m$