Question

A stone of mass $2 \, kg$ is projected upward with the kinetic energy of $98 \, J$ . The height at which the kinetic energy of the stone becomes half of its original value is given by

• A. $5m$
• B. $2.5m$
• C. $1.5 \, m$
• D. $0.5 \, m$

Answer 1

Answer: (B)

At the time of projection kinetic energy of the stone,
$ K = \frac{1}{2} \textit{mu}^{2}$
where m is the mass of the stone and u is the velocity of the projection
where $m$ is the mass of the stone and $u$ is the velocity of the
projection
or $u^{2}=\frac{2 K}{m}=\frac{2 \times 98}{2}=98$
$U \sin g, v^{2}=u^{2}-2 g h$
$\therefore \quad h=\frac{u^{2}}{2 g } \quad(\because \quad v =0)$
$h=\frac{98}{2 \times 9.8}=5 m$
Also, $\quad K=\frac{1}{2} m(2 gh ) \quad$ (Using
$K^{\prime}=\frac{1}{2} m v ^{\prime 2}=\frac{1}{2} m \times\left(2 gh ^{\prime}\right)$
$\therefore \quad \frac{K^{\prime}}{K}=\frac{h^{\prime}}{h}$
According to the problem
$K^{\prime}=\frac{K}{2}$
$\frac{K}{2 K}=\frac{h^{\prime}}{h}$
$h^{\prime}=\frac{h}{2}=\frac{5}{2} \quad m =2 \cdot 5 m$