Answer:
29.396988 m/s
Explanation:
Really, it depends on where the child is when he drops the ball - e.g., which planet he is on, and his distance from the center of that planet.
I'll assume that the child is on Earth at sea level at the equator, so that his distance from the geocenter is 6378000 meters.
The acceleration, g, is found from
g = GM/r²
G = 6.6743e-11 m³ kg⁻¹ sec⁻²
M = 5.9724e+24 kg
r = 6.378e+6 m
g = 9.799086 m sec⁻²
An approximate answer is found from an equation from constant acceleration kinematics:
v = gt
t = 3.0 sec
v = 29.397259 m/s
Now, the above method is an approximation that makes the technically incorrect assumption that the acceleration of gravity is a constant throughout the entire fall. You get away with it because the drop is very short. In another situation, it might not be. So it would be nice to develop a more accurate method that does not assume constant gravitational acceleration. For that, we begin with the Vis Viva equation:
v = √[GM(2/r − 1/a)]
Here,
a = the semimajor axis of a plunge orbit, which is equal to half of the apoapsis distance of 6378000+h, where
h = the altitude from which the ball is dropped
We can (using some math) develop the following equation:
t − t₀ = √[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }
t − t₀ = 3 sec
r = 6378000 meters
d = r + h
Using an iterative method (e.g. Newton's or Danby's), we can determine that the altitude,
h = 44.0954 meters
So,
d = 6378044.09538 meters
a = d/2 = 3189022.04769 meters
Now we can calculate that
v = 29.396988 m/s
This is the more nearly correct answer because it takes into account the variability of the gravitational acceleration during the fall.
and an angle of 
. On the y-axis (vertical direction), it is an accelerated motion with acceleration equal to -g (gravitational acceleration). The vertical velocity of the projectile at any time t is given by