Figure 2-45 shows a simple device for measuring your reaction time. It consists of a cardboard strip marked with a scale and two large dots. A friend holds the strip vertically, with thumb and forefinger at the dot on the right in Fig. 2-45. You then position your thumb and forefinger at the other dot (on the left in Fig. 2-45), being careful not to touch the strip. Your friend releases the strip, and you try to pinch it as soon as possible after you see it begin to fall. The mark at the place where you pinch the strip gives your reaction time. (a) How far from the lower dot should you place the 50.0 m smark? How much higher should you place the marks for (b) 100, (c) 150, (d) 200, and (e) 250 ms? (For example, should the 100 msmarker be 2 times as far from the dot as the 50msmarker? If so, give an answer of 2 times. Can you find any pattern in the answers?

The problem deals with the kinematic equation of motion. Also, it involves the concept of free fall. Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence motion. The equations which are used in the study are known as kinematic equations of motion. The situation of a body moving freely in any direction in the presence of gravity is referred to as free fall.

Formula:

The displacement in kinematic equation is given by,

$∆y=\left(v_0\right)\left(t\right)+\frac{1}{2}a{t}^2$

Where, $v_0$is the initial velocity a is acceleration and t is time.

According to the kinematic equations,

$$\begin{gathered} ∆y=\left(v_0\right)\left(t\right)+\frac{1}{2}a{t}^2 \\ 0-y_1=0-\left(\frac{1}{2}\right)\left(9.8\right){\left(0.05\right)}^2 \\ {y}_1=4.9\left(0.0025\right) \\ {y}_1=0.0123\mathrm{m} \\ ~1.23\mathrm{cm} \end{gathered}$$

Hence, the distance of lower dot from mark would be 1.23 cm

$$\begin{gathered} ∆y=\left(v_0\right)\left(t\right)+\frac{1}{2}a{t}^2 \\ 0-y_2=0-\left(\frac{1}{2}\right)\left(9.8\right){\left(0.1\right)}^2 \\ {y}_2=4.9\left(0.01\right) \\ {y}_2=0.049\mathrm{m} \\ =4{\mathrm{y}}_1 \end{gathered}$$

Hence, the distance of lower dot from 100 ms mark would be$4y_1$

$$\begin{gathered} ∆y=\left(v_0\right)\left(t\right)+\frac{1}{2}a{t}^2 \\ 0-y_3=0-\left(\frac{1}{2}\right)\left(9.8\right){\left(0.15\right)}^2 \\ {y}_3=4.9\left(0.0225\right) \\ {y}_3=0.11\mathrm{m} \\ =9y_1 \end{gathered}$$

Hence, the distance of lower dot from 150 ms mark would be$9y_1$

$$\begin{gathered} ∆y=\left(v_0\right)\left(t\right)+\frac{1}{2}a{t}^2 \\ 0-y_4=0-\left(\frac{1}{2}\right)\left(9.8\right){\left(0.2\right)}^2 \\ {y}_4=4.9\left(0.04\right) \\ {y}_4=0.196\mathrm{m} \\ =16y_1 \end{gathered}$$

Hence, the distance of lower dot from 200 ms mark would be$16y_1$

$$\begin{gathered} ∆y=\left(v_0\right)\left(t\right)+\frac{1}{2}a{t}^2 \\ 0-y_5=0-\left(\frac{1}{2}\right)\left(9.8\right){\left(0.25\right)}^2 \\ {y}_5=4.9\left(0.0625\right) \\ {y}_5=0.306\mathrm{m} \\ =25y_1 \end{gathered}$$

Hence, the distance of lower dot from 250 ms mark would be$25y_1$