Answer
The camera has fallen 13.2 meters after 4.0 seconds.
Work Step by Step
We can find the acceleration:
$v_f = v_0+at$
$a = \frac{v_f-v_0}{t}$
$a = \frac{3.3~m/s-0}{2.0~s}$
$a = 1.65~m/s^2$
We can find the distance the camera falls after $4.0~s$:
$\Delta y = \frac{1}{2}at^2$
$\Delta y = \frac{1}{2}(1.65~m/s^2)(4.0~s)^2$
$\Delta y = 13.2~m$
The camera has fallen 13.2 meters after 4.0 seconds.
