Question: you throw a ball. assume that the origin is on the ground, with the$+y$ axis pointing upward. Just after the ball leaves your hand its position is$\left(0.06,1.03,0\right)m$ . The average velocity of the ball over the next $0.7s$is $\left(17,4,6\right)m$. As time $0.7s$after the ball leaves your hand, what is the height of the ball above the ground?

The given data can be listed below as,

• The location of the ball is$\left(0.06,1.030\right)m$ .
• The average velocity of the ball is$\left(17,4,6\right)m/s$ .
• The time taken for reaching the average velocity is$0.7s$ .

This law illustrates that a particular body will continue to move in a uniform motion unless an external force acts upon that particle.

The equation of displacement gives the height of the ball above the ground.

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 is and

$t=$The time taken by the ball.

Substituting the values localid="1657704657139" $s_0=\left(0.06,1.03,0\right)m$, $v=\left(17,4,6\right)m/s$, and $t=0.7s$ in the above equation we get:

$$\begin{gathered} s=\left(0.06,1.03,0\right)m+\left(0.7s\right)\times \left(\left(17,4,6\right)m/s\right) \\ s=\left(0.06,1.03,0\right)m+\left(11.9,2.8,4.2\right)m.s/s \\ s=\left(11.96,3.83,4.2\right)m \end{gathered}$$

Thus, the height of the ball above the ground is 3.83m.