A diver makes 2.5 revolution 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

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\frac{5 π}{\sqrt{2}} rad / s$

$\frac{π}{\sqrt{2}} rad / s$

$\frac{3 π}{\sqrt{2}} rad / s$

$\frac{5 π}{\sqrt{3}} rad / s$

answer is B.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The free-fall time:

$∆ y = v_{o} t + \frac{1}{2} {gt}^{2} \Rightarrow t = \sqrt{\frac{2 (10 m)}{10 m / s^{2}}} = \sqrt{2} s$

Thus, the magnitude of the average angular velocity is

$ω_{avg} = \frac{(2.5 rev) (2 π rad / dev)}{\sqrt{2} s} = \frac{5 π}{\sqrt{2}} rad / s$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test