A mass of $10\ kg$ is dropped from a height of $50\ cm$. Find its : (i) Kinetic energy (ii) Velocity just as it reaches the ground. [Take $g=10\ ms^{-2}$]


As given, initial velocity $u=0$

mass $m=10\ kg$

Distance travelled $=$ Height $h=50\ cm=\frac{50}{100}\ m=\frac{1}{2}\ m$

Acceleration $g=10\ ms^{-2}$

Using the equation of motion $v^2=u^2+2gh$

We have $v^2=0+2\times10\times\frac{1}{2}$

Or $v^2=10\ ms^{-1}$

Therefore kinetic energy $K=\frac{1}{2}mv^2$

Or $K=\frac{1}{2}\times10\times10=50\ Joule$

Thus, $( i)$ Kinetic energy of the particle is $50\ Joule$ and $(ii)$ velocity of the particle just as it reaches the ground is $10\ ms^{-1}$.

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Updated on: 10-Oct-2022

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