How High Does the Faulty Model Rocket Reach?

In summary, the problem involves a faulty model rocket with acceleration components given by ax= \alphat^2 and ay=\beta-\gammat , where \alpha=2.50m/s^4, \beta=9.00m/s^2, and \gamma=1.40m/s^3. The initial velocity of the rocket is \overline{v0} = v0x\widehat{i} + v0y\widehat{j}with v0x = 1m/s and v0y = 7m/s. The goal is to find the maximum height reached by the rocket. To solve this, the quadratic equation is used with the y component of equation 2 to find the
  • #1
clope023
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Homework Statement



A faulty model rocket moves in the xy-plane (the positive y-direction is vertically upward). The rocket's acceleration has components ax= [tex]\alpha[/tex]t^2 and ay=[tex]\beta[/tex]-[tex]\gamma[/tex]t , where [tex]\alpha[/tex]=2.50m/s^4, [tex]\beta[/tex]=9.00m/s^2, and [tex]\gamma[/tex]=1.40m/s^3. At t=0 the rocket is at the origin and has velocity [tex]\overline{v0}[/tex] = v0x[tex]\widehat{i}[/tex] + v0y[tex]\widehat{j}[/tex]with v0x = 1m/s and v0y = 7m/s.

question: what is the rockets' maximum height reached?

Homework Equations



1. [tex]\overline{v}[/tex](t) = v0x + [tex]\alpha[/tex]t^3/3, v0y + ([tex]\beta[/tex]t-[tex]\gamma[/tex]t^2/2)

2. [tex]\overline{r}[/tex](t) = v0xt + [tex]\alpha[/tex]t^4/12, v0yt + ([tex]\beta[/tex]t^2/2 - [tex]\gamma[/tex]t^3/6)

other equations I tried, might help

3. x = (v0cos[tex]\alpha[/tex]0)t

4. y = (v0sin[tex]\alpha[/tex]0)t-1/2gt^2

5. vx = v0cos[tex]\alpha[/tex]0

6. vy = v0sin[tex]\alpha[/tex]0-gt

7. y-y0 = v0yt+1/2gt^2

8. quadratic equation -b = +/- [tex]\sqrt{b^2-4ac}[/tex]/2a

The Attempt at a Solution



this is a masteringphysics problem in which I was supposed to originally find the velocity and position vectors which easily enough I found to be the integral of the acceleration and the integral of velocity respectively.

I originally tried using the quadratic equation with the components of the y component of equation 2 to solve for time and plug that time back into the y component of the position vector and solve for y.

problem was I kept on getting very tiny numbers and eventually I ran out of attempts and the masteringphysics system just gave the answer to be 341m, I have no idea how they did that with the info presented, if anyone could give me any hints as to how the question is done that would be very helpful.
 
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  • #2
I understand that the maximum height is when the velocity vector is 0, so I know that at the maximum height the y component of the velocity vector must be 0 which I believe is why they used the quadratic equation. I also tried using equation 4 with the components of equation 1 and plugging in the t value from the quadratic equation into equation 5 but I still get a tiny number for the maximum height. Any help would be greatly appreciated, thanks in advance!
 
  • #3




It seems like you have attempted to solve the problem correctly using the equations provided. However, it is possible that there may be a mistake in your calculations or in the values given for the acceleration components. I would suggest double-checking your work and also trying the problem using different methods or equations to see if you get the same answer. It is also helpful to draw a diagram of the motion and label all known values to better visualize the problem. If you are still having trouble, you could also consult with your instructor or a classmate for assistance.
 

1. What is 2D motion?

2D motion refers to motion that occurs in two dimensions, typically represented on a flat surface such as a graph or a coordinate plane. In the case of a model rocket, this would involve the rocket's motion in the vertical (up and down) and horizontal (side to side) directions.

2. How is 2D motion different from 3D motion?

Unlike 2D motion, which occurs in two dimensions, 3D motion occurs in three dimensions, including the vertical, horizontal, and depth directions. This means that a model rocket's motion in 3D would also take into account any changes in its depth or distance from the launch point.

3. What factors affect the 2D motion of a model rocket?

The 2D motion of a model rocket can be affected by various factors, including its initial velocity, air resistance, gravitational force, and any external forces acting on the rocket (such as wind or thrust from the rocket engine). Additionally, the angle of launch and the weight and shape of the rocket can also impact its 2D motion.

4. How can I calculate the 2D motion of a model rocket?

The 2D motion of a model rocket can be calculated using mathematical equations, such as the kinematic equations for projectile motion. These equations take into account the initial velocity, acceleration, and time of flight of the rocket to determine its position and velocity at any given time.

5. What are some practical applications of studying the 2D motion of model rockets?

Studying the 2D motion of model rockets can have various practical applications, such as improving the design and performance of rockets used in space exploration or missile technology. It can also be used to teach physics concepts and principles, such as projectile motion and Newton's laws of motion, to students and enthusiasts interested in rocketry and aerospace engineering.

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