A diver makes `2.5` revolutions on the way from a `10-m-high` platform to the water. Assuming zero initial vertical velocity, the average angular velocity during the dive is.

Options

A. `(3 pi)/(sqrt(2)) rad//s`
B. `(5 pi)/(sqrt(2)) rad//s`
C. `(5 pi)/(sqrt(3)) rad//s`
D. `(pi)/(sqrt(2)) rad//s`

Answer

B

Explanation

(b) The free-fall time :
`Deltay = v_(0 y) t +(1)/(2) gt^2 rArr t = sqrt((2(10 m))/(10 m//s^2)) = sqrt(2) s`
Thus, the magnitude of the average angular velocity is.
`omega_(avg) =((2.5 rev)(2 pi rad//rev))/(sqrt(2) s) =(5 pi)/(sqrt(2)) rad//s`.