Question

A 13 kg stone drops from a height of 17m. Neglecting air resistance, what will be its velocity just before it hits the ground? Use the law of conservation of energy to solve this problem, NOT an equation of motion. g= 9.81 m/s².

Answer 1

$$PE=KE,$$ $$mgh=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{2gh}=\sqrt{2\cdot9.81\ \dfrac{m}{s^2}\cdot17\ m}=18.26\ \dfrac{m}{s^2}.$$