A man drops a $10\ kg$ rock from the top of a $5\ m$ ladder. What is its speed just before it hits the ground? What is its kinetic energy when it reaches the ground? $(g=10\ m/s^2)$

Here given, mass of the rock $m=10\ kg$

Height of the ladder $h=5\ m$

Gravitational acceleration $g=10\ m/s^2$

On using the third equation of motion $v^2=u^2+2gh$

Or $v^2=0+2\times10\ m/s^2\times 5\ m$

Or $v^2=100\ m^2/s^2$

Or $v=10\ m/s$

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

$=\frac{1}{2}\times 10\ kg.\times100\ m^2/s^2$

$=500\ Joule$

Therefore, the kinetic energy of the rock when it reaches the ground is $500\ J$.

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Updated on: 10-Oct-2022

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