If an armadillo can leap to 0.5m height, how fast must he be moving vertically when his feet just lift off the pavement?

To determine the speed at which the armadillo must be moving vertically when its feet just lift off the pavement, we can make use of the principles of physics.

The armadillo's vertical velocity when its feet leave the ground can be determined using the equation:

v^2 = u^2 + 2as

Where:
v = final vertical velocity (which is 0 since the armadillo's feet just lift off the pavement)
u = initial vertical velocity
a = acceleration
s = vertical displacement

In this case, as the armadillo is moving vertically, the acceleration due to gravity (a) will be acting downwards. Therefore, its value will be -9.8 m/s^2 (negative because it is in the opposite direction of the positive vertical motion).

The vertical displacement (s) is the maximum height the armadillo can leap, so s = 0.5 m.

Plugging in the values into the equation:

0 = u^2 + 2(-9.8)(0.5)

Simplifying the equation:

0 = u^2 - 9.8

Rearranging the equation to solve for u:

u^2 = 9.8

Taking the square root of both sides:

u = √9.8

Calculating the value of u:

u ≈ 3.13 m/s

Therefore, the armadillo must be moving vertically upwards with a speed of approximately 3.13 m/s when its feet just lift off the pavement.