Question

Question: Two diamonds begin a free fall from rest from the same height, $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$ apart. How long after the first diamond begins to fall will the two diamonds be $\mathbf{10}\mathbf{}\mathbf{m}$ apart?

Short answer

After $1.5\mathrm{s}$, the two diamonds would be $10\mathrm{m}$apart.

Step-by-step solution

Given information

$∆\mathrm{y}=10\mathrm{m}$

Initial velocity$\left({\mathrm{v}}_0\right)=0$

To understand the concept

The problem deals with the kinematic equations of motion. Kinematics is the study 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,

$∆y=v_{0}t+\frac{1}{2}a{t}^2$

Calculation for time

Location of the first diamond is given by

$$\begin{gathered} \mathrm{y}=0\times \mathrm{t}+\frac{1}{2}\times \mathrm{g}\times {\mathrm{t}}^2 \\ \Rightarrow \mathrm{y}=\frac{1}{2}\times \mathrm{g}\times {\mathrm{t}}^2...............\left(\mathrm{i}\right) \end{gathered}$$

Location of the second diamond is given by

$$\begin{gathered} \mathrm{y}=0\times \mathrm{t}+\frac{1}{2}\times \mathrm{g}\times {\mathrm{t}}^2 \\ \Rightarrow {\mathrm{y}}_0=\frac{1}{2}\times \mathrm{g}\times {\left(\mathrm{t}-1\right)}^2...............\left(\mathrm{ii}\right) \end{gathered}$$
$$\begin{gathered} {\mathrm{y}}_0=0\times \mathrm{t}+\frac{1}{2}\times \mathrm{g}\times {\mathrm{t}}^2 \\ \Rightarrow {\mathrm{y}}_0=\frac{1}{2}\times \mathrm{g}\times {\left(\mathrm{t}-1\right)}^2 \end{gathered}$$

Now, the difference between the two locations should be $10\mathrm{m}$. So, subtracting the second

$$\begin{gathered} \mathrm{y}-{\mathrm{y}}_0=\frac{1}{2}\times \mathrm{g}\times {\mathrm{t}}^2-\frac{1}{2}\times \mathrm{g}\times {\left(\mathrm{t}-1\right)}^2 \\ 10=\frac{1}{2}\times \mathrm{g}\times {\mathrm{t}}^2-\frac{1}{2}\times \mathrm{g}\times {\left(\mathrm{t}-1\right)}^2 \\ \frac{20}{\mathrm{g}}={\mathrm{t}}^2-{\left(\mathrm{t}-1\right)}^2 \\ \frac{20}{9.8}={\mathrm{t}}^2-{\left(\mathrm{t}-1\right)}^2 \\ {\mathrm{t}}^2-\mathrm{t}+2\mathrm{t}-1=2.04 \\ 2\mathrm{t}=3.04 \\ \mathrm{t}=1.5\mathrm{s} \end{gathered}$$

Hence, the time at which the two diamonds are $10\mathrm{m}$apart from each other would be $1.5\mathrm{s}$