Answer:
x = 8.74m
Explanation:
Hello! Let's solve this!
Given a mass m = 0.44kg
v = 4m / s
F = 8N
y = 20m
Force on the rocket
F '= F * - (mg) = m * a
We cleared to
a = (F * (- mg)) / m = (F / m) * (- g)
The rocket will accelerate upwards at a time t given by x = v * t
We cleared t
t = x / v
From kinematics we have to
y = (1/2) * a * t2
We replace a and t
y = (1/2) * (F / m) * (- g) * (x / v) 2
We clear x that is the incognita.
[tex]x=\sqrt{(2*y*v^{2} )/(F/m)-g}[/tex]
[tex]x=\sqrt{(2*20m*(4)^{2} )/((8N/0.44kg)-9.8m/s^{2} )}[/tex]
x = 8.74m
The horizontal distance to launch the rocket from is 8.74 meters.
Given the following data:
Scientific data:
From Newton's Second Law of Motion, the net force exerted on the rocket is given by:
[tex]F_n=F-mg=ma\\\\a=\frac{F-mg}{m} \\\\a=\frac{-Fg}{m}[/tex]
From the second equation of motion (kinematics), the horizontal distance to launch the rocket from is given by this formula:
[tex]y = ut+\frac{1}{2} at^2 \\\\y = 0(t)+\frac{1}{2} (\frac{-Fg}{m} )t^2\\\\y = \frac{1}{2} (\frac{-Fg}{m} )(\frac{x}{v} )^2\\\\x=\sqrt{\frac{2yv^2}{\frac{-Fg}{m}} } \\\\x=\sqrt{\frac{2myv^2}{-Fg} } \\\\x=\sqrt{\frac{2 \times 0.44 \times 20 \times 4^2}{-8 \times 9.8} }[/tex]
x = 8.74 meters.
Read more on acceleration here: brainly.com/question/24728358
