91Ó°ÊÓ

When an item is released from rest in the open air (with no friction), its speed is entirely determined by gravity's acceleration, g.

If we drop an item of mass m from a height h near the Earth's surface, it has initial mechanical energy as:

$\mathbf{U}\mathbf{=}\mathbf{mgh}$

When an item collides with the ground, all mechanical energy (potential energy) is converted to kinetic energy as:

$\mathbf{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$

Where v denotes the speed of the object shortly before it collides with the earth.

If we know what speed v is safe for the object's integrity, we can determine how high it was dropped.

As described, kinetic energy will be equal to potential energy for the case.

After equating the equation, we can find out the value of h.

This value of height is the safe height.

$\begin{array}{c}\mathrm{mgh}=\frac{1}{2}{\mathrm{mv}}^2\\ \mathrm{h}=\frac{{\mathrm{mv}}^2}{2\mathrm{mg}}\\ \mathrm{h}=\frac{{\mathrm{v}}^2}{2\mathrm{g}}\end{array}$

If the case happened on the moon, then the value would have become,

${\mathrm{h}}_{\mathrm{M}}=\frac{{{\mathrm{v}}_{\mathrm{M}}}^2}{2{\mathrm{g}}_{\mathrm{M}}}$

Now it is very clear that the gravitational acceleration in the moon is 1/6^th times that of the gravitational pull of the earth.

Hence if we put that value of g_m in the equation, we get,

$\begin{array}{c}{\mathrm{h}}_{\mathrm{M}}=\frac{{\mathrm{v}}^2}{2×\left(\frac{1}{6}\mathrm{g}\right)}\\ {\mathrm{h}}_{\mathrm{M}}=6\left(\frac{{\mathrm{v}}^2}{2×\left(\mathrm{g}\right)}\right)\\ {\mathrm{h}}_{\mathrm{M}}=6\mathrm{h}\end{array}$

Hence it is clear that an object could fall from six times the original height and still be safe.