How can I calculate the maximum height reached by a model rocket?

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To determine the maximum height of a model rocket that takes 5 seconds to return to the launch site, the key is to understand the effects of gravity on its ascent and descent. The formula for distance under constant acceleration, s(t) = 1/2 a t^2, is applicable here, with acceleration due to gravity set at -9.8 m/s^2. The rocket's peak height can be calculated by considering the average speed during its fall, which is derived from its initial speed of 0 m/s and final speed after 5 seconds. This results in a maximum height of approximately 122.5 meters, confirming the answer of 120 meters when rounded. Understanding the principles of constant acceleration is crucial for solving such physics problems.
hlcfairy
I have to take a final soon and I have practise questions The only one here I am having trouble with is thislast one. This question bugs me and the answer key doesn't show how to do it

After a model rocket reached its maximum height, it then too 5.0 seconds to return to the launch site. What is the approximate maximum height reachd by the rocket? {Neglect air resistance.}

the answer is 120 m

the only info u have is time and acceleration

How do I go about doing it?
 
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the rocket going up is confusing, but i think this is what you're looking for:
simply, D=RT (distance=rate x time, if you're not familiar with it)

my guess is that this is not correct becuase you have to take into account the return of the rocket because of gravity. so something you muliply by either 2 or 1/2. maybe...
 
Last edited:
The question asks: how far does something fall in 5 seconds, when its acceleration (due to gravity) is a = 9.8 m/s^2?

s(t) = 1/2 a t2, where t = 5, yields 122.5 m.

- Warren
 
Maximus, "D= RT" only applies when R is a constant.

If R is changing at a constant rate (that is the rate of change of R is a constant) then you can use an "averaging" method. When the rocket was at it's peak, it's speed was 0 (that's why it stopped going up!). After 5 seconds at a constant acceleration of -9.8 m/s^2, it's speed is -9.8*5= -49 m/s (Since the rate of change of speed, acceleration, is constant, you CAN use "RT"). The average of the two values is (0+(-49)/2= -24.5. Using that average value, in 5 seconds, the rocket will fall -24.5*5= -122.5 m. The rocket must have fallen from a height of 122.5 m, value chroot (Warren) gave.

Using "g" instead of -9.8 m/s^2 and t instead of 5 seconds, the two speeds are 0 and -gt so the "average" speed is (0-gt)/2= -gt/2.
Multiplying that by t to get distance gives (-g/2)t^2, the formula chroot used.

Again, this only works when the acceleration is constant (a very important special case!). If the acceleration (in general "rate of change of the rate of change") is not constant, then you will have to learn calculus!
 
I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

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