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

`y=7+1.73x-0.0016x^(2)`

Parbolic path.

Explanation:

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

`y=y_(0)+(tg(\theta))x-(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=(g)/(2(v_(0)cos(\theta))^(2))=0.0016 (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!