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Brums [2.3K]
3 years ago
10

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
Mathematics
1 answer:
Soloha48 [4]3 years ago
6 0

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!

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