At a certain instant, a ball passes location $\left(7,21,-17\right)\mathbf{}\mathit{m}$.In the next $\mathbf{3}\mathbf{}\mathit{s}$,the ball’s average velocity is$\left(-11,42,-11\right)\mathit{m}\mathbf{/}\mathit{s}$. At the end of this $\mathbf{3}\mathbf{}\mathit{s}$ time interval, what is the height y of the ball?

The given data can be listed below as

• The location of the ball is $\left(7,21,-17\right)\mathit{m}$.

• The average velocity of the ball is $\left(-11,42,-11\right)\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$.

The time taken for reaching the average velocity is $3s$.

This law states that a body will continue to be in rest or uniform motion unless an external force acts upon the body.

The equation of displacement gives the height of the ball.

From Newton’s first law, the equation of displacement for the ball is expressed as:

$s=s_0+vt$

Here, $s=$The height of the ball,

$s_0=$The initial displacement of the ball,

$v=$The average velocity of the ball

$t=$ The time taken by the ball.

Substituting the values ${\mathit{s}}_{\mathbf{0}}\mathbf{=}\left(7,21,-17\right)\mathit{m}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}\mathit{v}\mathbf{=}\left(11,42,-11\right)\mathit{m}\mathbf{/}\mathit{s}$and $\mathit{t}\mathbf{=}\mathbf{3}\mathit{s}$in the above equation, we get:

$$\begin{gathered} \mathit{s}\mathbf{=}\left(7,21,-17\right)\mathit{m}\mathbf{+}\left(3s\right)\mathbf{\times }\left(\left(11,42,-11\right)m/s\right) \\ \mathit{s}\mathbf{=}\left(7,21,-17\right)\mathit{m}\mathbf{+}\left(33,126,-33\right)\mathit{m} \\ \mathit{s}\mathbf{=}\left(40,147,-50\right)\mathit{m} \end{gathered}$$

Thus, the height y of the ball is $\mathbf{147}\mathit{m}$.