College Physics (4th Edition)

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.