Physics
posted by James on .
By trial and error, a frog learns that it can leap a maximum horizontal distance of 1.4 m. If, in the course of an hour, the frog spends 32% of the time resting and 68% of the time performing identical jumps of that maximum length, in a straight line, what is the distance traveled by the frog?

Yes, it's a super frog.

The 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.