91Ó°ÊÓ

The given data can be listed below,

• The value of constant is,$\mathrm{Î\pm }=2.50m/s^4$
• The value of the constant for y-component is,$\mathrm{β}=9.0m/s^2$
• The initial x-component of velocity of rocket is,$v_{0x}=1.0m/s$
• The initial y-component of velocity of rocket is,$v_{0y}=7.0m/s$

An object's position may be described using its position vector. It is essential to understand a body's location in order to accurately describe its motion.

The velocity of the rocket is given by integrating acceleration which is given as,

${∫}_{v_0}^{v}dv={∫}_{v_0}^{v}adt$

Substitute all the values in the above,

The x-component of the velocity is calculated as,

$\begin{array}{rcl}{v}_x-v_{0x}& =& {∫}_0^t\mathrm{Î\pm }{t}^{2}dt\\ v_x& =& v_{0x}+\frac{1}{3}\mathrm{Î\pm }{t}^2\end{array}$

The y-componentsof the rocket is given by,

$\begin{array}{rcl}{∫}_{v_0}^{v}d{v}_y& =& {∫}_0^t\left(\mathrm{β}-\mathrm{γ}t\right)dt\\ v_y-v_{0y}& =& {\overline{)\left(\mathrm{β}t-\frac{1}{2}\mathrm{γ}{t}^2\right)}}_0^t\\ v_y& =& v_{0y}+\mathrm{β}t-\mathrm{γ}{t}^2\end{array}$

The velocity vector of the rocket is given by,

role="math" localid="1659684055862" $\stackrel{â‡\P Ä}{v}\left(t\right)=v_{0x}+\frac{1}{3}\mathrm{Î\pm }{t}^3+v_{0y}+\mathrm{β}t-\mathrm{γ}{t}^2$

The x-component of the position is given by,

$\begin{array}{rcl}{v}_x& =& \frac{dx}{dt}\\ x-x_0& =& {∫}_0^t{v}_{x}dt\end{array}$

Substitute all the values in the above,

role="math" localid="1659683975821" $\begin{array}{rcl}x-x_0& =& {∫}_0^t{v}_{0x}+\frac{1}{3}\mathrm{Î\pm }{t}^{3}dt\\ & =& v_{ox}t+\frac{1}{12}\mathrm{Î\pm }{t}^4\\ x& =& x_0+v_{0x}t+\frac{1}{12}\mathrm{Î\pm }{t}^4\end{array}$

The y-component of position is given by,

$$\begin{gathered} y-y_0={∫}_0^t\left(v_{0y}+\mathrm{β}t-\frac{1}{2}\mathrm{γ}{t}^2\right)dt \\ =v_{oy}t+\frac{1}{2}\mathrm{β}{t}^2-\frac{1}{6}\mathrm{γ}{t}^3 \\ y=y_0+v_{0y}t+\frac{1}{2}\mathrm{β}{t}^2-\frac{1}{6}\mathrm{γ}{t}^3 \end{gathered}$$

Thus, the position of the rocket is$\begin{array}{rcl}x& =& x_0+v_{0x}t+\frac{1}{12}\mathrm{Î\pm }{t}^4\end{array}$ and $y=y_0+v_{0y}t+\frac{1}{2}\mathrm{β}{t}^2-\frac{1}{6}\mathrm{γ}{t}^3$, the velocity as a function time is $v_{0x}+\frac{1}{3}\mathrm{Î\pm }{t}^3+v_{0y}+\mathrm{β}t-\mathrm{γ}{t}^2$.

At the maximum height, the vertical component of velocity is zero. The heights of the rocket is given by,

$\begin{array}{rcl}{v}_0& =& v_{0y}+\mathrm{β}t-\frac{1}{2}\mathrm{γ}{t}^2\\ 0& =& v_{0y}+\mathrm{β}t-\frac{1}{2}\mathrm{γ}{t}^2\end{array}$

Substitute all the values in the above,

$\begin{array}{rcl}0& =& 7m/s+\left(9m/s^2\right)t-\frac{1}{2}\left(1.40m/s^2\right)t^2\\ t& =& 13.6s\end{array}$

Substitute all the values in the equation for y-component of position is given by,

$\begin{array}{rcl}y& =& 7×13.6+\frac{9×{\left(13.6\right)}^2}{2}-\frac{1.4×{\left(13.6\right)}^3}{6}\\ & =& 340.6m\end{array}$

Thus, the maximum height of the rocket is 340.6 m

The time of the rocket by the equations given by,

$\begin{array}{rcl}{v}_{0y}t+\frac{1}{2}\mathrm{β}{t}^2-\frac{1}{6}\mathrm{γ}{t}^3& =& 0\\ \frac{1}{6}\mathrm{γ}{t}^2-\frac{1}{2}\mathrm{β}t-v_{0y}& =& 0\end{array}$

Substitute all the values in the above,

$\begin{array}{rcl}\left(1.40\right)\frac{1}{6}{t}^2-\frac{1}{2}\left(9.0\right)t-v_{0y}& =& 0\\ t& =& 20.73s\end{array}$

Put the value of time in the equation for x-component of position,$\begin{array}{rcl}x& =& 1×20.73+\frac{2.5×{\left(20.73\right)}^4}{12}\\ & =& 38500m\\ & =& 3.85×{10}^{4}m\end{array}$

The graph for trajectory is shown below as,

From graph it is clear that the x and y components are not symmetric.

Thus, the horizontal displacement of the rocket is 3.85x10⁴ m