To solve this problem, we can use conservation of energy. When the egg falls from the nest, it has gravitational potential energy (due to its height) that is gradually converted to kinetic energy (due to its motion) as it falls.
Using conservation of energy, we can equate the initial potential energy of the egg to the final kinetic energy just before it hits the ground:
mgh = (1/2)mv^2
where m is the mass of the egg, g is the acceleration due to gravity (9.81 m/s^2), h is the initial height of the egg (3.0 m), and v is the speed of the egg just before it hits the ground.
Solving for v, we get:
v = sqrt(2gh)
Plugging in the given values, we get:
v = sqrt(2 x 9.81 m/s^2 x 3.0 m) = 7.67 m/s
So the speed of the egg when it is 0.50 m from the ground is approximately 7.67 m/s.