Question

A child slides down the water slide at a swimming pool and enters the water at a final speed of 4.22 m/s. At what final speed would the child enter the water if the water slide were twice as high? Ignore friction and resistance from the air and the water lubricating the slide.

Answer 1

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:

$mgh = \frac {1}{2} mv^2\\v =\sqrt{2gh}$

let the final speed be v' when h' = 2 h

$\frac{v'}{v}=\frac{\sqrt h'}{\sqrt h}\\v'=\sqrt{\frac{2h}{h}}\times v\\v'=\sqrt2 v\\v'=5.97 m/s$