The path of a model rocket

In summary, the problem involves a model rocket being launched with an initial velocity of 50.0 m/s and accelerating at a constant rate of 2.00 m/s^2 until reaching an altitude of 150 m. The question asks for the time it takes for the rocket to reach its maximum height. Two different methods are used to solve the problem, with one giving the correct answer of 8.52 seconds and the other giving an incorrect answer of 5.49 seconds.
  • #1
professordad
18
0

Homework Statement



A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s^2 until its engines stop at an altitude of 150 m. How long after lift off does the rocket reach the maximum height?

from College Physics by Serway and Faughn

Homework Equations



(1) [itex]v = v_0 + at[/itex] where [itex]v[/itex] denotes velocity, [itex]v_0[/itex] denotes initial velocity, [itex]a[/itex] denotes acceleration, and [itex]t[/itex] denotes time in seconds.

(2) [itex]x = v_0t + \frac{1}{2}at^2[/itex] where [itex]x[/itex] denotes the displacement

(3) [itex]v^2 = v_0^2 + 2ax[/itex]

The Attempt at a Solution



Ok so for this problem I found 2 ways to do it. One of them gives the correct answer (in the back of the book) and the other is off by a little.

The method that gives the correct answer:


So let the velocity of the rocket at the point where the acceleration stops be [itex]v_1[/itex] and the velocity of the rocket at the highest point be [itex]v_2[/itex]. Let the time of the rocket's acceleration be [itex]t_1[/itex] and the time of no rocket's acceleration be [itex]t_2[/itex]. Note that at the highest point, the velocity is 0, so [itex]v_2 = 0[/itex].

Now we find [itex]v_1[/itex]. From equation (3), [itex]v_1^2 - 50.0^2 = 2 \cdot 2.00 \cdot 150[/itex] so [itex]v_1 = 55.68[/itex] m/s.

From equation (1), we have [itex]v_1 = 50.0 + at_1[/itex], so [itex]55.68 = 50.0 + 2.00 \cdot t_1[/itex] and [itex]t_1 = 2.84[/itex] sec.

From equation (1) again, we have [itex]v_2 = v_1 + at_2[/itex], so [itex]-55.68 = -9.8t[/itex] and [itex]t_2 = 5.68[/itex] sec.

The answer is [itex]\boxed{8.52}[/itex] sec.


Ok so here's the other method that I tried but it didn't give the right answer:

Let [itex]d_2[/itex] denote the distance that the rocket travels after it stops accelerating itself. Then, by (3), [itex]0 = 55.68^2 - 2 \cdot 9.8 \cdot x[/itex], so [itex]x = 158[/itex].

So by (2), [itex]x = 158 = 55.68 \cdot t_2 - 4.9t_2^2[/itex].

Solving the quadratic gives [itex]t_2 = 5.49[/itex] sec or [itex]t = 5.87[/itex] sec.

But this doesn't give the right answer :( , help?
 
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  • #2
I get t=√3100/9.8=5.681 sec.
 
  • #3
Yes :) because [itex]\sqrt{3100} = 55.68[/itex] as obtained in my first method, but why doesn't the second method work?
 
  • #4
professordad said:
1. .


Ok so here's the other method that I tried but it didn't give the right answer:

Let [itex]d_2[/itex] denote the distance that the rocket travels after it stops accelerating itself. Then, by (3), [itex]0 = 55.68^2 - 2 \cdot 9.8 \cdot x[/itex], so [itex]x = 158[/itex].

So by (2), [itex]x = 158 = 55.68 \cdot t_2 - 4.9t_2^2[/itex].

Solving the quadratic gives [itex]t_2 = 5.49[/itex] sec or [itex]t = 5.87[/itex] sec.

But this doesn't give the right answer :( , help?


I also wonder why 2 different values for t.
But if we put displacement=0 then,

0=55.68-4.9t
ttotal=55.68/4.9=11.3632
t=5.68163
 
Last edited:
  • #5




Both of the methods you have used are correct, but the second method may have a slight error in the calculations. It is always important to double check your work and make sure all calculations are correct. Additionally, make sure to use consistent units throughout your calculations (e.g. converting all units to meters and seconds). This could also be a possible source of error. It is also helpful to label all quantities in your equations to avoid confusion. Overall, your approach to solving the problem is correct and with careful calculations, you should be able to arrive at the correct answer.
 
  • #6




Thank you for sharing your work and thought process for solving this problem. Your first method is correct and gives the correct answer. However, your second method has a small mistake in the equation for distance traveled after the rocket stops accelerating. It should be x = 150 = 55.68t - 4.9t^2, which gives t = 5.87 sec as you calculated. However, since the rocket reaches its maximum height at an altitude of 150 m, the correct equation should be x = 150 = 55.68t - 4.9t^2 - 150, which gives t = 8.52 sec, the same answer as your first method. Keep up the good work in solving problems and remember to double check your equations for accuracy.
 

1. What factors affect the path of a model rocket?

The path of a model rocket is primarily affected by three factors: thrust, gravity, and air resistance. Thrust is the force that propels the rocket upward, while gravity pulls it back down. Air resistance, also known as drag, can slow down the rocket's ascent and change its trajectory.

2. How does the shape of a model rocket affect its flight path?

The shape of a model rocket directly impacts its flight path. A streamlined design with a pointed nose cone and fins can reduce air resistance and help the rocket fly straighter. On the other hand, a bulky or unevenly shaped rocket may experience more drag and have a less predictable path.

3. What role do weather conditions play in the path of a model rocket?

Weather conditions can significantly affect the path of a model rocket. Wind, in particular, can cause the rocket to veer off course, especially if it is not aerodynamically stable. Rain or high humidity can also add extra weight to the rocket, making it more difficult to reach its intended altitude.

4. How does the weight of a model rocket affect its trajectory?

The weight of a model rocket can impact its trajectory in two ways. First, a heavier rocket may require more thrust to overcome gravity and reach higher altitudes. Second, a heavier rocket may experience more air resistance, causing it to slow down and change direction more easily.

5. What are some common safety concerns when launching a model rocket?

When launching a model rocket, it is essential to follow safety guidelines to ensure the safety of yourself and others. Some common safety concerns include launching in a clear, open area, avoiding power lines and trees, and using recovery systems, such as parachutes, to safely bring the rocket back down to the ground.

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