91Ó°ÊÓ

A body accelerates when a net force acts on it and travels in the same direction as the net force. The product of mass and acceleration vector is equal to the force vector.

$∑\stackrel{→}{F}=m\stackrel{→}{a}$

For constant acceleration, the final velocity v is related to the initial velocity u by the expression given below,

$\mathit{v}\mathbf{=}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}$

The force acted on the body at the end of the fall will depend on the de-acceleration and the time it took to stop the fall completely.

$$\begin{gathered} F=ma \\ =m\left(\frac{v-u}{t}\right) \\ =m\left(\frac{0-u}{t}\right) \\ =-\frac{mu}{t} \end{gathered}$$

The final velocity in both cases will be the same. But the concrete street offers high de-acceleration, i.e., the time twill be really short. As a result, the person will be seriously injured at the bottom. In water, the person can go inside the water where de-acceleration is quite small; hence the time t is long. The person will stop at some depth in time t which will not result in any injury.

Hence, it can be said that It’s not the $\mathit{v}\mathbf{=}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}$that hurts you; it’s the sudden change in the momentum, i.e $\mathit{F}\mathbf{=}\mathbf{-}\frac{\mathbf{m}\mathbf{u}}{\mathbf{t}}$at the bottom.