91Ó°ÊÓ

The kinematic equation mainly defines motion of a particle with respect to the time. The kinematic equation gives the distance travelled by an object that is the division between the difference between the square of final and initial velocity and the two times of the acceleration.

The equation of the distance of the small steel ball is expressed as:

$v_f^2=v_i^2+2g\mathrm{Δ}y$

Here,$v_f$ is the final speed, is the initial speed, $v_i$is the distance travelled by the steel ball and$\mathrm{Δ}y$ is the acceleration due to gravity.

As initially the ball was at rest, then the initial velocity of the ball is zero.

Substitute 0 for v in the above equation.

$$\begin{gathered} v_f^2=2g\mathrm{Δ}y \\ \mathrm{Δ}y=\frac{v_f^2}{2g} \\ =\frac{1}{2g}v \end{gathered}$$

The above equation is the equation of the straight line having the slope $\frac{1}{2g}$.

The equation of the slope is expressed as:

$s=\frac{1}{2g}$

Here,g is the acceleration due to gravity with value 9.8 m/s².

Substitute 9.8m/s² for g in the above equation.

$$\begin{gathered} s=\frac{1}{2\left(9.8{\text{m/s}}^{\text{2}}\right)} \\ =\frac{1}{\left(19.6{\text{m/s}}^{\text{2}}\right)} \\ =0.0510{\text{s}}^{\text{2}}\text{/m} \end{gathered}$$

Thus, the numerical value of the slope of this straight line is $0.0510{\text{s}}^{\text{2}}\text{/m}$.

The acceleration is less than the value 9.8 m/s² of when the air resistance is present. Hence, as the acceleration is less, then the ball’s final speed will also be less.

Thus, the final speed for a given release height is lower when I ignored the air resistance.

When the air resistance is present, then there will not be any constant acceleration. Hence, the graph will not behave like a straight line. The final velocity will be less in the presence of the air resistance.

The graph of$∆y$ versus$v_f^2$ has been provided below:

According to the graph, the line is not a straight line. Hence, with the increase in the final velocity, the distance also increases.