Answer:
v'=5.97 m/s
Explanation:
If the friction and fluid friction are ignored, then by law of conservation of mechanical energy, potential energy at the top of the slide must be equal to the kinetic energy at the bottom of the slide. Thus, the height of slide and final speed at the bottom of the slide are related as:
[tex]mgh = \frac {1}{2} mv^2\\v =\sqrt{2gh}[/tex]
let the final speed be v' when h' = 2 h
[tex]\frac{v'}{v}=\frac{\sqrt h'}{\sqrt h}\\v'=\sqrt{\frac{2h}{h}}\times v\\v'=\sqrt2 v\\v'=5.97 m/s[/tex]
