Question

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

• 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.
$v_{1}^{2}=u^{2}-2 g h_{1}$
$\therefore (10)^{2}=u^{2}-2 \times 10 \frac{h}{2}$
$\left(\because h_{1}=\frac{h}{2}, v_{1}=10 ms ^{-1}\right)$
or $100=u^{2}-10\, h$ ...(i)
Again at height $h$
$v_{2}^{2} =u^{2}-2 g h$
$0=u^{2}-2 \times 10 \times h \left(\because v_{2}=0\right)$
$0=u^{2}-20\, h$ ...(ii)
Subtracting Eq. (ii) from Eq. (i), we get
or $100 =10\, h$
$h=\frac{100}{10}=10\, m$