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April 18, 2014

April 18, 2014

Posted by **James** on Monday, May 21, 2012 at 3:40pm.

- Physics -
**James**, Monday, May 21, 2012 at 3:40pmYes, it's a super frog.

- Physics -
**Elena**, Monday, May 21, 2012 at 4:12pmThe time of jumping is

0.68•3600 = 2448 sec.

The distance covered by this small “projectile” (the frog) during 1 jump is

L =v²•sin2α/g.

We have Lmax => sin2α =1 and α = 45º.

Lmax= v² /g.

v =sqrt(Lmax•g) = sqrt(1.4•9.8) =3.7 m/s.

The time of 1 jump is

t = 2•v•sin α/g = 2•3.7•sin45º/9.8 =0.53 sec.

During the hour the frog made N jumps:

N = 2448/0.53=4619.

The total distance is

1.4•4619 =6467 m.

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