91Ó°ÊÓ

$$\begin{gathered} 1)\mathrm{Time}\mathrm{interval}\mathrm{is}\mathrm{t}=∆{\mathrm{t}}_1 \\ 2)\mathrm{Initial}\mathrm{velocity},{\mathrm{V}}_{0\mathrm{x}}=0\mathrm{and}{\mathrm{V}}_{\mathrm{oy}}=0 \\ 3)\mathrm{The}\mathrm{old}\mathrm{coordinates}\mathrm{are}:\mathrm{A}=\left(4.00\mathrm{m},6.00\mathrm{m}\right)\mathrm{and}\mathrm{B}\left(12.0\mathrm{m},18.0\mathrm{m}\right) \end{gathered}$$

Projectile motion is the motion of a particle that is launched with an initial velocity. During its flight, the particle’s horizontal acceleration is zero and its vertical acceleration is the free-fall acceleration. (Upward is taken to be a positive direction.)

Using the projectile motion equations for motion along x and motion along the y-direction, we can calculate the ratio of acceleration components.

Formulae:

The second horizontal distance equation due to kinematics motion,

$\mathrm{x}-{\mathrm{x}}_0={\mathrm{V}}_{\mathrm{ox}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{x}}{\mathrm{t}}^2$ …(¾±)

The second vertical distance equation due to kinematics motion,

$\mathrm{y}-{\mathrm{y}}_0={\mathrm{V}}_{0\mathrm{y}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{y}}{\mathrm{t}}^2$ …(¾±¾±)

By using the given values in equation (i), we get the equation as:

$$\begin{gathered} \left(12\mathrm{m}-4\mathrm{m}\right)=0+\frac{1}{2}{\mathrm{a}}_{\mathrm{x}}∆{\mathrm{t}}_1^2 \\ 8\mathrm{m}=\frac{1}{2}{\mathrm{a}}_{\mathrm{x}}∆{\mathrm{t}}_1^2 \end{gathered}$$ …(¾±¾±¾±)

Similarly using the same data in equation (ii), we get the equation as:

$$\begin{gathered} \left(18\mathrm{m}-6\mathrm{m}\right)=0+\frac{1}{2}{\mathrm{a}}_{\mathrm{y}}∆{\mathrm{t}}_1^2 \\ 12\mathrm{m}=\frac{1}{2}{\mathrm{a}}_{\mathrm{y}}∆{\mathrm{t}}_1^2 \end{gathered}$$ …(¾±±¹)

Now, dividing eq. (iv) by eq. (iii), we get the required value of the ratio of the vertical to the horizontal accelerations as follows:

$$\begin{gathered} \frac{{\mathrm{a}}_{\mathrm{y}}}{{\mathrm{a}}_{\mathrm{x}}}=\frac{12.0\mathrm{m}}{8.00\mathrm{m}} \\ =1.5 \end{gathered}$$

Hence, the value of the ratio is 1.5.

It is given that the motion continues for additional time interval which is same as

For this part, we have the time interval as:$∆{\mathrm{t}}_2=2∆{\mathrm{t}}_1$

Substituting the given values in equation (i) we have

$8m=\frac{1}{2}{a}_x∆t_1^2$

Now for new displacement we can write the same equation (i) as follows:

$∆x=\frac{1}{2}{a}_x∆t_2^2$

Taking the ratio of these two above equations, we get the-coordinate as:

role="math" localid="1657017179060" $$\begin{gathered} \frac{∆x}{8m}=\frac{∆t_2^2}{∆t_1^2} \\ ∆x=\left(8.00m\right)×2^2\left(âˆ\mu ∆t_2=2∆t_1\right) \\ =32m \end{gathered}$$

Similarly, using the given data in equation (ii), we get the-coordinate as:

$$\begin{gathered} ∆\mathrm{y}=\left(12.0\mathrm{m}\right)×2^2 \\ =48\mathrm{m} \end{gathered}$$

So, the new co-ordinates will be, (considering the previous distance)

$$\begin{gathered} \mathrm{x}=32\mathrm{m}+4\mathrm{m} \\ =36\mathrm{m} \\ \mathrm{y}=48\mathrm{m}+6\mathrm{m} \\ =54\mathrm{m} \end{gathered}$$

Hence, the value of the new coordinates of the particle is$\left(36\mathrm{m},54\mathrm{m}\right)$