The problem is this: 5. A diver makes 2.5 revolutions on the way from a 10-m-hig

Question

The problem is this:

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

A Cramster expert answered this question and the answer made by the expert is here:

"Height of the platform ?x = -10 m
Acceleration a = -9.8 m/s
The time taken by the diver to reach the bottom of the platform is obtained by using
?x = 1/2 at2
t = v(2x/a)
= 1. 429 s
Therefore angular velocity during dive is
(2.5 rev/ 1.429s)(2p rad/ 1 rev) = 10.99 rad/s."

Please explain how 1.429 s is obtained in more detail. Thank you very much.

Explanation / Answer

We have,

s = ut + (1/2)at^2

s = (1/2)at^2

t = (2s/a)^(1/2)

s = 10 m ; a = g = 9.8 m/s^2

t = (20/9.8)^(1/2) = 1.429 s

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