Question

A student is “accidentally" pushed by Mr. M from the top of NHS. The roof of NHS to the ground (by the
cafeteria) is roughly 15.6 meters. If it takes 1.80 seconds to fall, what is the student's acceleration?

Answer 1

Answer:

$9.6 m/s^2$ downward

Explanation:

The motion of the student is a uniformly accelerated motion, so we can use the suvat equation

$s=ut+\frac{1}{2}at^2$

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:

$a=\frac{2s}{t^2}=\frac{2(15.6)}{1.80^2}=9.6 m/s^2$

And since the sign is positive, the direction of the acceleration is downward.