Do you remember this formula for the distance traveled while accelerated ?
Distance = (initial speed) x (t) plus (1/2) x (acceleration) x (t²)
I think this is exactly what we need for this problem.
initial speed = 20 m/s down
acceleration = 9.81 m/s² down
t = 3.0 seconds
Distance down = (20) x (3) plus (1/2) x (9.81) x (3)²
Distance = (60) plus (4.905) x (9)
Distance = (60) plus (44.145) = 104.145 meters
Choice D) is the closest one.
