A ball is thrown vertically downward from the top of a $\mathbf{36}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{m}$-tall building. The ball passes the top of a window that is$\mathbf{12}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{m}$above the ground$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{s}$after being thrown. What is the speed of the ball as it passes the top of the window? top speed. What must this distance be if he is to achieve a time of$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$for the race?

$\begin{array}{rcl}y& =& 36.6\text{m}\\ y_0& =& 12.2\text{m}\\ t& =& \text{2}\text{s}\end{array}$

The problem deals with the kinematic equations of motion. Kinematics is the studies of how a system of bodies moves without taking into account the forces or potential fields that influence the motion. The equations which are used in the study are known as kinematic equations of motion.

Formula:

The displacement is given by,

$\Delta y=v_{0}t+\frac{1}{2}a{t}^2$

The final velocity is given by,

$v_f^2=v_0^2+2a\Delta y$

Assume that the downward direction is positive and the origin is at the ground.

To find the speed use following equation

$y-y_0=v_i\times t+\frac{1}{2}\times a\times t^2$

$36.6-12.2=v_0\times 2+\frac{1}{2}\times 9.81\times 2^2$

$24.4=v_0\times 2+19.6$

$v_0=2.4\mathrm{m}/\mathrm{s}$

This is the initial speed with which stone was thrown.

Now using another kinematic equation

$v_f^2=v_0^2+2a\Delta y$

$v_f^2=2.4^2+2\times 9.8\times 24.4$

$v_f^2=484$

$v_f=22\text{m/s}$

So, the velocity would be$22\text{m/s}$