Answer:
[tex]9.6 m/s^2[/tex] downward
Explanation:
The motion of the student is a uniformly accelerated motion, so we can use the suvat equation
[tex]s=ut+\frac{1}{2}at^2[/tex]
where, taking downward as positive direction, we have
s = 15.6 m is the vertical displacement
u = 0 is the initial velocity (we assume the student starts from rest)
t = 1.80 s is the time of the fall
a is the acceleration of the student
By re-arranging the equation and solving for a, we can find the acceleration of the student:
[tex]a=\frac{2s}{t^2}=\frac{2(15.6)}{1.80^2}=9.6 m/s^2[/tex]
And since the sign is positive, the direction of the acceleration is downward.