Upward and downward movement of a rocket with the equation of motion

In summary: The statement "since it was at its peak it wasn't moving anymore so we can assume that s(t) and s0(t) = 0" means that the velocity is zero at that point.
  • #1
arhzz
266
52
Homework Statement
The equation of motion applies
A rocket starts at a speed of 42 m / s.
(a) How far did she fly after 1 s?
(b) What is the maximum height? When does the rocket reach its maximum altitude? ¨
(c) At what speed and when does the rocket hit the earth again?
(d) What initial speed would have been necessary for the rocket to take 20 s ¨
hits back on earth?
(e) Show that the rocket for the upward movement and for the downward movement ¨
takes the same length regardless of the initial speed.
Relevant Equations
s(t) = s0 + v0t - g/t^2
Hello! I've done the following to solve this problem.

a) Here I simply put in the time in the equation, s0 is = 0 and after that it was pretty much done

$$s(t) = 42 *1 - \frac {9,81*1^2} {2} = 37,09m $$

b) Now here to see when the rocket reaches it maximum altitude and what height it is, first we need to check at what point did the rocket reach its maximum altitude. Since it was at its peak it wasnt moving anymore so we can assume that s(t) and s0(t) = 0 and if we input that into the equation and try to get t out.

$$ 0 = 42 * \frac {9,81*t^2} {2} $$ Now we multiply by 2 get t to the left and t should be $$ t= 4,28 s$$

c) Now here for the velocity I simply assumed that it is also 42 m/s, actually -42 m/s. Because after it reaches it highest point its speed is 0. It will than have to travel the same distance to reach the ground. So the time and speed should be the same, simply in the other direction. That said the entire time the rocket was in the air should be $$ t_{total} = t1+t2 = 8,56 s $$

d) Now here is where I'm not getting a few things. So the time is 20 so t= 20. Now to get the v0 I simply rearange the equation of motion.

$$ s(t) = v0t - \frac{gt^2}{2} $$ Multiply by 2

$$ 2s = 2v0t - g *t^2$$ Now the 2 on the right an the left should cancel out so

$$ s = vot - g* t^2 $$ Now since we want v0 we move it to the left and we move s to the right side of the equation

$$ v0 = \frac {g*t^2} {t*s} $$

Now when I plug in all the values I get v0 = 5,28 m/s and that is obviously wrong. Now we are given the correct solution it is 98,1 m/s but I cannot get it. The steps that they took to get the solution are as following;

$$2s = v0t - gt^2 $$

$$ v0 = \frac {gt} {2} $$

Now I have no idea how u get from the first equation to this. If anyone understands this I'd be grateful if he could give me a litle bit of insight.

Thanks!
 
Physics news on Phys.org
  • #2
arhzz said:
d) Now here is where I'm not getting a few things. So the time is 20 so t= 20. Now to get the v0 I simply rearange the equation of motion.

$$ s(t) = v0t - \frac{gt^2}{2} $$ Multiply by 2

$$ 2s = 2v0t - g *t^2$$ Now the 2 on the right an the left should cancel out so

$$ s = vot - g* t^2 $$ Now since we want v0 we move it to the left and we move s to the right side of the equation
That's an interesting piece of algebra. The first and third equations are clearly not equivalent.
 
  • #3
I thought as much that the problem was in the way I went around with the formula. Thing is I am not sure how to do it right.
 
  • #4
arhzz said:
I thought as much that the problem was in the way I went around with the formula. Thing is I am not sure how to do it right.
You start with your equation: $$s(t) = v_0 t - \frac 1 2 gt^2$$ And you want to find ##v_0## such that when ##t = 20s##, we have ...?
 
  • #5
Um I am not quite sure I get what you are trying to say, but I will try nontheless. I want to find v0 such, that when the t = 20 we have a hit the Earth with the rocket. Which should mean that the distance traveled at that time point is 0?
 
  • #6
arhzz said:
Um I am not quite sure I get what you are trying to say, but I will try nontheless. I want to find v0 such, that when the t = 20 we have a hit the Earth with the rocket. Which should mean that the distance traveled at that time point is 0?
Yes, except that the displacement is zero: ##s(t) = 0##. Not the distance travelled.

So, what can you say about ##v_0## if ##s(20s) = 0##?

Or, more generally, if ##s(t) = 0## and ##t \ne 0##.
 
  • #7
Well I can say that s doesn't play a role in the equation itself, so the equation should be without displacement? I think you have given me a very good push in the right direction, I'll try solving it now.
 
  • #8
arhzz said:
b) ... Since it was at its peak it wasn't moving anymore so we can assume that s(t) and s0(t) = 0
##0 = 42 * \frac {9,81*t^2} {2}##
Not moving (instantaneously) is a statement about velocity, not position, so how do you get s(t)=0? (And what does s0(t) even mean?)
The equation above makes no sense. It would clearly lead to t=0. Yet by some magic you got the right answer for the time to peak height.
What I presume you solved was 0=42-gt.
I don't see an answer for max height.

In the most common form of SUVAT there are five variables and five equations. Each equation uses a different four of the five variables. So the way to use them is first to identify which variable you don't care about.
In (d), you know g and t, and you want to find v0. You need one more known; the choice is s or v1.
You seem to have opted for s, but not indicated what value you have for it.
 
  • #9
Okay for b) you are right I forgot to put in the value of max height s(4,28) = 89,91m.I forgot actually how I did this problem(the a b and c part) so I did it out of my head while typing here.I rechecked my calculations and found out that I actually used the equation for velocity here (v = v0 -gt). And v should be 0 assuming when we reach max altitude the velocity is 0. (Sorry for the missconfusion was rushing to get to part d)

Now for d) I'm still stuck here, I've been trying for about an hour and I'm not getting any further. Since we have g and t and need v0 so we choose from v1 or s. Now since I choose s, I'd assume the value would be 0? Am I interpreting this wrong? If I would to chose v1 that would be the speed, but how would I calculate the speed since I actually ned the v0 to get v? Or am I just flat out wrong?

Thanks!
 
  • #10
arhzz said:
Now since I choose s, I'd assume the value would be 0?
Yes, but you did not show that in post #1. Instead, you showed two errors:
- the mishandling of the factor of 2 that @PeroK pointed out in post #2.
- what I now take to be a typo:
##v0 = \frac {g*t^2} {t*s}##
I assume you meant, using s=0, ##v0 = \frac {g*t^2} {t}##
arhzz said:
If I would to chose v1 that would be the speed, but how would I calculate the speed since I actually need the v0 to get v?
What is the relationship between ##v_0## and ##v_1## in this case?
 
  • #11
Okay I actually did misstype the equation when using s=0, so that formula for v0 is wrong? Now the relationshop of v0 and v1 in this case, should be that v0 = v1 ?
 
  • #12
arhzz said:
Okay I actually did misstype the equation when using s=0, so that formula for v0 is wrong? Now the relationshop of v0 and v1 in this case, should be that v0 = v1 ?
You seem to be completely lost here. What you need to do is a very common general approach to such problems. Perhaps you haven't seen anything like this before. We start with your equation: $$s(t) = v_0 t - \frac 1 2 gt^2$$ And, this is a key point, we set ##s(t) = 0## to represent the time(s) when the particle is on the ground. This gives us the equation:$$0 = v_0 t - \frac 1 2 gt^2$$ which we will rewrite as $$v_0 t = \frac 1 2 gt^2$$ Now, we know ##t = 0## is a solution (the starting point). What we want is to look for a solution with ##t \ne 0##, which allows us to cancel a ##t## from both sides and get: $$v_0 = \frac 1 2 gt$$ This is a new equation for the relationship between the initial velocity and the total time of flight. This allows us to do two things: if we know the initial velocity, we can calculate the time of flight. And, if we know the time of flight, we can calculate the initial velocity.
 
  • Like
Likes arhzz
  • #13
arhzz said:
$$ s = vot - g* t^2 $$ Now since we want v0 we move it to the left and we move s to the right side of the equation
$$ v0 = \frac {g*t^2} {t*s} $$
:
Okay I actually did misstype the equation when using s=0, so that formula for v0 is wrong?
Of the two above equations, the first is wrong as @PeroK pointed out. You have lost a 2 somewhere. You had started with
$$ s = vot - \frac 12g* t^2 $$.

You have also gone wrong in getting from the first of the two above equations to the second. If you cannot see the error, please post the detailed steps.

I get the impression you are a bit error prone in manipulating equations. If so, you might find it helpful to take smaller steps.
arhzz said:
Now the relationshop of v0 and v1 in this case, should be that v0 = v1 ?
Not quite. There is an important difference. They are velocities, not speeds.
 
Last edited:
  • #14
I get it now thanks to both of you!
 

Related to Upward and downward movement of a rocket with the equation of motion

1. What is the equation of motion for the upward and downward movement of a rocket?

The equation of motion for the upward and downward movement of a rocket is given by the second law of motion, F=ma, where F is the net force acting on the rocket, m is the mass of the rocket, and a is the acceleration of the rocket.

2. How does the mass of the rocket affect its upward and downward movement?

The mass of the rocket affects its upward and downward movement because according to the equation of motion, F=ma, the acceleration of the rocket is directly proportional to its mass. This means that a heavier rocket will require a greater force to accelerate it, and therefore, it will have a slower upward and downward movement compared to a lighter rocket.

3. What factors can affect the net force acting on a rocket during its upward and downward movement?

The net force acting on a rocket during its upward and downward movement can be affected by various factors such as the thrust of the rocket engines, air resistance, and gravity. The thrust of the engines provides the force needed to overcome the rocket's weight and lift it off the ground. Air resistance, on the other hand, acts in the opposite direction of the rocket's movement and can slow it down. Gravity also plays a significant role as it pulls the rocket towards the ground, causing it to accelerate downwards.

4. How does air resistance affect the upward and downward movement of a rocket?

Air resistance can significantly affect the upward and downward movement of a rocket. As the rocket moves through the air, it experiences a force in the opposite direction of its movement due to air resistance. This force can slow down the rocket's speed and affect its trajectory. Therefore, it is essential for rockets to be designed to minimize air resistance to achieve efficient and stable upward and downward movement.

5. Can the equation of motion be used to predict the exact path of a rocket during its upward and downward movement?

No, the equation of motion alone cannot predict the exact path of a rocket during its upward and downward movement. This is because there are many other factors that can affect the rocket's trajectory, such as wind, air density, and external forces. However, the equation of motion can provide a general understanding of the forces acting on the rocket and how they affect its movement.

Similar threads

  • Introductory Physics Homework Help
Replies
14
Views
838
  • Introductory Physics Homework Help
Replies
2
Views
283
  • Introductory Physics Homework Help
Replies
19
Views
720
  • Introductory Physics Homework Help
Replies
13
Views
829
  • Introductory Physics Homework Help
Replies
6
Views
771
  • Introductory Physics Homework Help
Replies
5
Views
816
  • Introductory Physics Homework Help
Replies
16
Views
456
  • Introductory Physics Homework Help
Replies
10
Views
733
  • Introductory Physics Homework Help
Replies
30
Views
585
  • Introductory Physics Homework Help
Replies
3
Views
188
Back
Top