Answer:
v = 12.28 m/s
Explanation:
Given that,
A stone is dropped from a height of 7.70 m.
We need to find the speed with which is it going when it reaches the ground. We can solve it using the conservation of energy as follows :
[tex]\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 7.7} \\\\v=12.28\ m/s[/tex]
So, it is moving with a speed of 12.28 m/s when it reaches the ground.
