**To calculate lift off velocity (of a jump that goes straight up) you can use the following formula:**
**mgh = 1/2(mV^2) where**

**m = mass of flea**

**g = acceleration due to gravity = 32.17 ft/s^2 on earth**

**h = max height of jump in feet**

**V^2 = take off velocity (in feet per second) squared**

**The expression on the left is the potential energy of the mass at the height, h.**

**The expression on the right of the equals sign is the kinetic energy of the mass at the velocity, V.**

**In a vacuum, or with big objects at relatively slow speeds, the
expressions can be equal. In other words, if you drop an object from an
h of 10 feet it will accelerate to a certain velocity after falling 10
feet. To jump 10 feet an animal of the same mass would have to be
traveling at the same velocity when it took off. potential energy at
max height of jump = kinetic energy at start of jump. 1st law of
thermodymics energy at start equals energy at finish if it doesn't go
somewhere else like into overcoming air resistance. **

**To solve the problem, note that the masses cancel out and you can have**

**V = (2gh)^1/2 Unfortunately I can't write math stuff well here.
The formula says the velocity at lift off equals the square root of (2
times g times h). If you use the units of feet for height and ft/sec
for g, then the velocity will be in feet per second. I'll leave it to
you to convert that to miles per hour or whatever units you are
comfortable with. **

**This formula applies only to an animal jumping straight up. If
the jumper goes off at an angle you have to know the angle and use a
little vector math. **

**From the moment the feet of the jumper leave the ground gravity starts to slow them down, so only take off velocity matters. **

**Unfortunately with tiny animals air resistance exerts such a huge
effect that the formula doesn't work for them. It works for horses and
people, probably even lemurs and rabbits, but not fleas. Trying to
calculate the varying drag force due to air resistance is quite a bit
more calculated and would require some experimentally determined
coefficients. Suffice it to say the actual take off velocity has to
be somewhat faster than would be required in a vacuum.**

** **