Question

A student is “accidentally" pushed by Mr. M from the top of NHS. The roof of NHS to the ground (by the

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.