Accepted Answer
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 downacceleration = 9.81 m/s² downt = 3.0 secondsDistance down = (20) x (3) plus (1/2) x (9.81) x (3)²Distance = (60) plus (4.905) x (9)Distance = (60) plus (44.145) = 104.145 metersChoice D) is the closest one.