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Unfortunately there is a frog sitting in the middle of the creek, who can jump with initial velocity 5m/s in either direction, and swallow the grasshopper in mid-air.

What is minimal initial velocity required to jump over the frog to the other bank of the creek?

2007-08-06 05:12:38 · 2 answers · asked by Alexander 6 in Science & Mathematics Physics

2 answers

We need to know the width of the creek. Lets say it is 'a' metres.

The maximum height the frog will acheive by jumping vertically upwards. So for the cricket to jump higher, the vertical component of its velocity must be greater than 5m/s.
For the horizontal component of the velocity it will be
v=d/t=a/t.
We get time 't' from the vertical motion:
s = ut - 0.5gt^2
Solve for 't' and plug in above equation.
Velocity will then be the square root of the the square sum of the components:
sqrt(horizontal velocity^2 + vertical velocity^2)

2007-08-06 06:48:00 · answer #1 · answered by qspeechc 4 · 0 0

Hey, what happened to my posted answer? Well, here it is again, in abbreviated form. The minimum velocity is 5√2, which will give the grasshopper exactly twice the maximum range as the frog's. No matter which direction and when the frog jumps, he can only barely touch the grasshopper's trajectory.

2007-08-06 08:25:57 · answer #2 · answered by Scythian1950 7 · 1 0

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