An object, initially at rest, falls from a bookshelf to the floor. If its final speed is 6 m/s, what is the height of the bookshelf?

Under the assumptions, the object would accelerate at a constant [tex]a = g = 9.81\; {\rm m\cdot s^{-2}}[/tex] towards the ground during the entire flight. It is also given that:

Initial speed is [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] since the object was initially at rest.
Speed right before landing is [tex]v = 6\; {\rm m\cdot s^{-1}}[/tex].
The distance travelled during the motion, which is the same as the height of the bookshelf, needs to be found. This quantity would be equal in magnitude to the displacement during the motion, [tex]x[/tex].

In this question, the duration [tex]t[/tex] of the motion is neither provided nor required. Hence, the SUVAT equation [tex](v^{2} - u^{2}) = 2\, a\, x[/tex] would be helpful for relating the other quantities about this motion: [tex]v[/tex], [tex]u[/tex], [tex]a[/tex], and [tex]x[/tex]. Rearrange this equation to express displacement [tex]x[/tex] in terms of [tex]v[/tex], [tex]u[/tex], and [tex]a[/tex]:

[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a} \\ &= \frac{(6\; {\rm m\cdot s^{-1}})^{2} - (0\; {\rm m\cdot s^{-1}})^{2}}{2\, (9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 0.31\; {\rm m} \end{aligned}[/tex].

In other words, the displacement of this object would be approximately [tex]0.31\; {\rm m}[/tex] (towards the ground) during the flight. The height of the bookshelf would be equal to the magnitude of this displacement: approximately [tex]0.31\; {\rm m}[/tex].