Answer:
His impact velocity is 7 m/s.
Explanation:
A stunt performer falls off a wall that is 2.5 m high and then lands on a mat. Let v is his impact velocity.
Concept used : Law of conservation of energy
Using the law of conservation of energy of the person such that,
[tex]\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh}[/tex]
g is acceleration due to gravity, g = 9.8 m/s²
[tex]v=\sqrt{2\times 9.8\times 2.5}\\\\v=7\ m/s[/tex]
So, his impact velocity is 7 m/s.
