Question

A rocket is launched from the top of a 7ft platform. Its initial velocity is 112 ft per sec. It is launched at an angle of​ 60° with respect to the ground. ​(a) Find the rectangular equation that models its path. What type of path does the rocket​ follow

Answer 1

Answer:

$y=7+1.73x-0.0016x^{2}$

Parbolic path.

Step-by-step explanation:

This is bidimensional motion, so the equation that relates the vertical and horizontal position is:

$y=y_{0}+(tg(\theta))x-\frac{g}{2(v_{0}cos(\theta))^{2}}x^{2}$

Here, v₀, θ y g are constants, then we can rewrite (1) as:

$y=a+bx-cx^{2}$

where:

• $a=y_{0}=7 ft$
• $b=tg(\theta)=1.73$
• $c=\frac{g}{2(v_{0}cos(\theta))^{2}}=0.0016 \frac{1}{ft}$  

Therefore the rectangular equation will be:

$y=7+1.73x-0.0016x^{2}$

This type of path is a parabolic motion.

I hope it helps you!