A rocket accelerates upwards Mechanics

[lubo]

Homework Statement

A rocket accelerates upwards from the ground at 25ms^-2 for 2.5 s at an angle of 80 degrees to the horizontal. The rocket motor then stops and eventually falls to the ground.

Neglecting air reststance and assuming thet the trajectory during acceleration is a straight line:

a. Make a labelled sketch of the rockets trajectory starting from the launch point. Include the numerical values of the horizontal and vertical velocity at the point where the rocket motor stops.

b. How long after launch does the rocket hit the ground?

c. How far does the rocket travel horizontally from the launch point?

Homework Equations

v=u²+2as
S=ut+1/2at²
Uh=xcosθ
Uv=xcosθ
v=u+at and v=u+gt

The Attempt at a Solution

See attached for diagram etc. I cannot actually upload it at this time, it will not lwt me? I can update it later if necessary.

v=u+at =0+25*2,5 = 62.5m/s

Uh = 62.5cos80 = 10.853 m/s
Uv = 62.5sin80 = 61.55 m/s

b.

linear distance traveled under acceleration = s=0*t+1/2*25*2.5² = 78.125m

This part is now for the point where there is no acceleration:

Uh = 62.5cos80 = 10.853 m/s
Uv = 62.5sin80 = 61.55 m/s

Now V has changed to U.

when V=0 at max height V²=U²+2S

Thus S = -62.5/2*-9.81 = 3.19m extra to the top with no acceleration now.

H1 = 78.125m + 3.19m = 81.315m

For time to max height:

V=U+gt = 0=-62.5*-9.81t Thus t2 = 6.37s

Now time to drop from max height:

-81.315 = 0*t+1/2*-9.81*t²

t=4s

t1 = 2.5 sec = acceleration phase.
t2 = 6.37 from after acceleration to the top
t3 = 4 sec time to fall back down

Total = 12.87

c. How far does the rocket travel:

S=Uh*t Uh = 10.853
S=Uh*2.5= 27.1325
S=Uh*6.37= 69.13
S=Uh*4= 43.41

S = 139.67.

I have done this and have no answers and do not know if it is right or where I might have gone wrong?

Any help is much appreciated and thanks.

 

[LawrenceC]

"when V=0 at max height V2=U2+2S

Thus S = -62.5/2*-9.81 = 3.19m extra to the top with no acceleration now."

62.5 m/s is the vector sum of velocity components. Use the vertical component of velocity.

V2 should be V^2. Equation needs a g before the S but you did insert it in the arithemetic.

 

[omoplata]

I think you've gone wrong at part b.

The distance the rocket will drop down to ground is the distance the rocket has traveled VERTICALLY, not the total distance the rocket has travelled.

Also, the gravitational acceleration acts vertically down, so when applying $V^2=U^2+2 a S$, you need to consider the VERTICAL VELOCITY of the rocket.

The same goes for finding the time t2.

Since your t2 and t3 are wrong, your answer for part c is also wrong. Otherwise it would be correct.

 

[lubo]

Thanks for your help, please see attached sheets for further confirmation, I am still unable to work this one out?

LawrenceC

you said that "62.5 m/s is the vector sum of velocity components. Use the vertical component of velocity."

Do you mean use:

Uv = 62.5sin80 = 61.55 m/s. Where?

"V2 should be V^2" thanks for that, my mis typing put in v=u2+2as but I know its v².

Omoplata

"I think you've gone wrong at part b.

"The distance the rocket will drop down to ground is the distance the rocket has traveled VERTICALLY, not the total distance the rocket has travelled."

I think I have gone wrong on Part B also, but i do not know how to rectify it? I know what you mean by the distance VERTICALLY, but I thought I had achieved the Horizontal distance at the end of the question?

 

[lubo]

Uploaded the documents as it had not worked.

 

[omoplata]

you said that "62.5 m/s is the vector sum of velocity components. Use the vertical component of velocity."

Do you mean use:

Uv = 62.5sin80 = 61.55 m/s. Where?

Yes, if you are using kinematic equations to find the distance the rocket has traveled vertically, you have to use that vertical component ( 61.55 m/s ) in the kinematic equations.

What you've tried to do is find the time and the distance the rocket traveled vertically until it stops, and then try to find the time it takes for the rocket to fall back down to the ground. But the distance you've found is not the vertical distance the rocket travelled.

I'll suggest a slightly different method to solve part b. You've tried to break the motion into three steps: the motion under accleration of 2.5 m/s^2, the motion under gravitational acceleration until the rocket stops, the motion under gravitational acceleration until the rocket hits the ground. But that is not necessary. You can break it into two steps: the motion under the acceleration of 2.5 m/s^2, the motion under gravity.

For the motion under the 2.5 m/s^2 acceleration, you know the acceleration, initial velocity, final velocity and the time of motion. To find the time it takes for the rocket to hit the ground under gravity, you need to find the distance traveled vertically in the first part. To do that, considering the numbers you already know, I suggest you use $v^2 = u^2 + 2 a S$ in one dimension (vertically), to find the vertical distance travelled, $S$. But be careful about the signs ( positive or negative ) and to use the vertical components of all the all the vectors here.

For the motion under gravity, you know the initial velocity, the acceleration and you've just found out the distance travelled. You can use $S = u t + \frac{1}{2} a t^2$ one dimensionally ( only vertically ) to find the time taken. Be careful about the signs and use only the vertical components of the vectors.

Once you find the time it took for the rocket to fall, you can add it to the time taken until the beginning of falling from the launch ( 2.5 s ), and you have the answer for part b.

"I think you've gone wrong at part b.

"The distance the rocket will drop down to ground is the distance the rocket has traveled VERTICALLY, not the total distance the rocket has travelled."

I think I have gone wrong on Part B also, but i do not know how to rectify it? I know what you mean by the distance VERTICALLY, but I thought I had achieved the Horizontal distance at the end of the question?

I hope the method outlined above for solving part b is clear. Let me know if it's not.

They ask for the HORIZONTAL distance in part c. To solve part c, also break the motion down to two parts: the motion under 2.5 m/s^2 acceleration, and the motion under gravity. For the fist part, use an equation like $S = u t + \frac{1}{2} a t^2$ or $v^2 = u^2 + 2 a S$ to find the distance traveled in that part. For the second part, the horizontal velocity doesn't change. So you can find the distance traveled in that part by multiplying the horizontal velocity by the time taken, which you found in part b. Add the two distances together, and you have the answer.

 

[SammyS]

When you used

v²=u²-2as ,​

where v is the velocity at the highest point of the trajectory, you can use either

the magnitudes of both u and v, in which case u = 62.5 m/s and v = 6.25 cos(80°) ≈ 10.853 m/s​

OR

the vertical components of both u and v, in which case, u_v = 62.5 sin(80°) ≈ 61.55 m/s and v_v = 0 .​

Also, t₃ cannot be less than t₂ .

 

[heartOFphysic]

Hi, sorry if the the following is completely wrong (though do point it out if it is wrong)...but what about this. Since acceleration is a vector it can be resolved into components.

so Vertical acceleration = 25 sin (80) = 24.62.

now we will work out the displacement of the rocket while it is going up (this is going to be the same when it comes down).

so a=24.62 v=0 t =2.5

so using v=u+at we get the initial vertical velocity as 61.55m/s. now s= (u+v)/2 t so this gives s= 76.94.

so now we have the rocket in the air coming down and we need to find how long it takes.

u=0 a=-9.8 s=76.94 t=? so s=1/2 at^2 can be applied to find t and 2.5 can be added to t to get the total time?

Is that right?

 

[omoplata]

Hi, sorry if the the following is completely wrong (though do point it out if it is wrong)...but what about this. Since acceleration is a vector it can be resolved into components.

so Vertical acceleration = 25 sin (80) = 24.62.

now we will work out the displacement of the rocket while it is going up (this is going to be the same when it comes down).

so a=24.62 v=0 t =2.5

The question says the rocket accelerates upwards. So I think it starts from rest and picks up speed. What we don't know is the final velocity. That is, we know that a=24.62, u =0 and t=2.5. But we don't know v.

so using v=u+at we get the initial vertical velocity as 61.55m/s. now s= (u+v)/2 t so this gives s= 76.94.

Maybe you did it correct but got the variables mixed up? Anyway, u=0 and v=61.55. The FINAL velocity is 61.55 m/s.

so now we have the rocket in the air coming down and we need to find how long it takes.

No, it's not coming down yet. It's still going up with a velocity of 61.55 m/s.

u=0 a=-9.8 s=76.94 t=? so s=1/2 at^2 can be applied to find t and 2.5 can be added to t to get the total time?

Is that right?

u = 61.55 , a = -9.8 s = -76.94 ( the rocket has to fall DOWN ), t = ?

You can use s = ut + (1/2)at^2 to find t, and by adding 2.5 can get the total time.

 

[lubo]

The Attempt at a Solution

v=u+at =0+25*2,5 = 62.5m/s

Uh = 62.5cos80 = 10.853 m/s
Uv = 62.5sin80 = 61.55 m/s[/COLOR]

So we now agree the above answer is correct.

b.

linear distance traveled under acceleration = s=0*t+1/2*25*2.5² = 78.125m

What about this?. Do I use this to find the distances fo acceleration part? i.e.

Uh = 78.125cos80 = 13.57 m S1h horizontal
Uv = 78.125sin80 = 76.94 m s1v (this would = the height of the acceleration part)?


This part is now for the point where there is no acceleration:

Uh = 62.5cos80 = 10.853 m/s
Uv = 62.5sin80 = 61.55 m/s

For the motion under the 2.5 m/s^2 acceleration, you know the acceleration, initial velocity, final velocity and the time of motion. To find the time it takes for the rocket to hit the ground under gravity, you need to find the distance traveled vertically in the first part. To do that, considering the numbers you already know, I suggest you use $v^2 = u^2 + 2 a S$ in one dimension (vertically), to find the vertical distance travelled, $S$. But be careful about the signs ( positive or negative ) and to use the vertical components of all the all the vectors here.

Now V has changed to U.

when V=0 at max height V²=U²+2S

Thus S = -61.55/2*-9.81 = 193.09m extra to the top with no acceleration now.

H1 = 76.94m + 193.09m = 270.03m

OR

For time to max height:

When you used

v²=u²-2as ,​

where v is the velocity at the highest point of the trajectory, you can use either

the magnitudes of both u and v, in which case u = 62.5 m/s and v = 6.25 cos(80°) ≈ 10.853 m/s​

OR

the vertical components of both u and v, in which case, u_v = 62.5 sin(80°) ≈ 61.55 m/s and v_v = 0 .​

V=U+gt = 0=-61.55*-9.81t Thus t2 = 6.27s


S = 61.55*t+1/2*-9.81*t² = 193m

H1 = 76.94m + 193.m = 269.94m TO THE TOP

do I need to find the Sh and Sv for 193m. I,e Sh = 193*cos80=33.53 and sv =190.18. ? The vector components? Please help me to understand this as this is where I AM FAILING. If it was a straight acceleration or gravity question i think i would be ok.

I DO NOT KNOW WHAT TO DO FROM HERE? SO SEE AS FOLLOWS:

For the motion under gravity, you know the initial velocity, the acceleration and you've just found out the distance travelled. You can use $S = u t + \frac{1}{2} a t^2$ one dimensionally ( only vertically ) to find the time taken. Be careful about the signs and use only the vertical components of the vectors.

S = 270.3m (Using this number)
g=-9.81
Uv= 0 or 61.22ms
t = ?

270.3 = 61.55t+0.5*-9.81*t²

As above this should be -3.443 or 16s.

OR IF Uv = 0 at the TOP THEN:

-270.3 = -4.905t²
t= √(-270.3/-4.905) = 55.1s

Once you find the time it took for the rocket to fall, you can add it to the time taken until the beginning of falling from the launch ( 2.5 s ), and you have the answer for part b.

?

If I take above I have 2.5s +16 = 18.5. I believe this is to the top.

OR 2.5 + 55.1 = 57.6 I think this is the correct way.



c. How far does the rocket travel:

They ask for the HORIZONTAL distance in part c. To solve part c, also break the motion down to two parts: the motion under 2.5 m/s^2 acceleration, and the motion under gravity. For the fist part, use an equation like $S = u t + \frac{1}{2} a t^2$ or $v^2 = u^2 + 2 a S$ to find the distance traveled in that part. For the second part, the horizontal velocity doesn't change. So you can find the distance traveled in that part by multiplying the horizontal velocity by the time taken, which you found in part b. Add the two distances together, and you have the answer.

acceleration = Uh = 78.125cos80 = 13.57 m S1h horizontal, if at the start I am correct.

Gravity = Uh = 62.5cos80 = 10.853 m/s * 55.1 = 598 m


any help is much appreciated and thanks again. This is very difficult for me and time consuming so any help to clear this up quick would be great.

 

[Doc Al]

The Attempt at a Solution

v=u+at =0+25*2,5 = 62.5m/s

Uh = 62.5cos80 = 10.853 m/s
Uv = 62.5sin80 = 61.55 m/s[/COLOR]

So we now agree the above answer is correct.

OK.

b.

linear distance traveled under acceleration = s=0*t+1/2*25*2.5² = 78.125m

What about this?. Do I use this to find the distances fo acceleration part? i.e.

Uh = 78.125cos80 = 13.57 m S1h horizontal
Uv = 78.125sin80 = 76.94 m s1v (this would = the height of the acceleration part)?

OK, that's the position of the rocket when the engines turn off (at the end of 2.5 seconds). (I don't understand why you call it U, since U usually stands for a velocity.)

This part is now for the point where there is no acceleration:

Uh = 62.5cos80 = 10.853 m/s
Uv = 62.5sin80 = 61.55 m/s

These are the initial velocity components for when it begins its motion as a free falling projectile. Not no acceleration! The acceleration is -g.

To find the time it takes to reach the ground, just analyze the vertical motion. What's the complete formula for position as a function of time? Then you can solve for the time it takes to reach the ground from that starting point.

 

[omoplata]

The Attempt at a Solution

v=u+at =0+25*2,5 = 62.5m/s

Uh = 62.5cos80 = 10.853 m/s
Uv = 62.5sin80 = 61.55 m/s[/COLOR]

So we now agree the above answer is correct.

Agreed.

b.

linear distance traveled under acceleration = s=0*t+1/2*25*2.5² = 78.125m

What about this?. Do I use this to find the distances fo acceleration part? i.e.

Uh = 78.125cos80 = 13.57 m S1h horizontal
Uv = 78.125sin80 = 76.94 m s1v (this would = the height of the acceleration part)?

Yes, you can do that.

This part is now for the point where there is no acceleration:

Uh = 62.5cos80 = 10.853 m/s
Uv = 62.5sin80 = 61.55 m/s




Now V has changed to U.

when V=0 at max height V²=U²+2S

Thus S = -61.55/2*-9.81 = 193.09m extra to the top with no acceleration now.

So, the answer is correct, but the intermediate steps should be like this,
$V^2 = U^2 + 2 a S$
$0 = 61.55^2 + 2 (-9.81) S$
$S = \frac{-(61.55^2)}{2 (-9.81)} = 193.09 m$

H1 = 76.94m + 193.09m = 270.03m

I agree.

OR

For time to max height:




V=U+gt = 0=-61.55*-9.81t Thus t2 = 6.27s

The answer is correct. But the intermediate steps should be,
$V = U + g t = 61.55 + (-9.81) t$
$t = 6.27$

S = 61.55*t+1/2*-9.81*t² = 193m

H1 = 76.94m + 193.m = 269.94m TO THE TOP

I agree.

do I need to find the Sh and Sv for 193m. I,e Sh = 193*cos80=33.53 and sv =190.18. ? The vector components? Please help me to understand this as this is where I AM FAILING. If it was a straight acceleration or gravity question i think i would be ok.

You don't need to find Sh and Sv for 193 m. What you found is Sv. That is, Sv = 193 m. The kinematic equations that you used were one dimensional. Since you used the vertical component of the initial velocity, what you get is the vertical component of the distance traveled.

I DO NOT KNOW WHAT TO DO FROM HERE? SO SEE AS FOLLOWS:



S = 270.3m (Using this number)
g=-9.81
Uv= 0 or 61.22ms
t = ?

Since you've already considered the distance the rocket travels under gravity until it stops, Uv = 0.

270.3 = 61.55t+0.5*-9.81*t²

As above this should be -3.443 or 16s.

No, Uv is not 61.55. Uv is 0.

OR IF Uv = 0 at the TOP THEN:

-270.3 = -4.905t²
t= √(-270.3/-4.905) = 55.1s




?

Actually, it should be,
$t = \sqrt{-270.3/-4.905} = 7.42$

If I take above I have 2.5s +16 = 18.5. I believe this is to the top.

OR 2.5 + 55.1 = 57.6 I think this is the correct way.

You've forgotten the 6.27 the rocket travels under gravity until rest.

So the total time traveled is 2.5 + 6.27 + 7.42.

c. How far does the rocket travel:




acceleration = Uh = 78.125cos80 = 13.57 m S1h horizontal, if at the start I am correct.

Gravity = Uh = 62.5cos80 = 10.853 m/s * 55.1 = 598 m

What does "Gravity = Uh" mean?

Anyway, the first 2.5 s is traveled under acceleration. So you have to consider that. The rest of the way is traveled under constant velocity.

 

[lubo]

OK.


OK, that's the position of the rocket when the engines turn off (at the end of 2.5 seconds). (I don't understand why you call it U, since U usually stands for a velocity).

I call it Uv as it is the initial vertical velocity. It is all I know?

To find the time it takes to reach the ground, just analyze the vertical motion. What's the complete formula for position as a function of time? Then you can solve for the time it takes to reach the ground from that starting point.

S=ut+1/2at²


omoplata: What does "Gravity = Uh" mean?
Anyway, the first 2.5 s is traveled under acceleration. So you have to consider that. The rest of the way is traveled under constant velocity.[/QUOTE]

EXPLINATION:

omoplata Thank you for the help and calculation checks.

It should be S2=Uh*t Therefore Uh = initial horizontal distance, but from the point the rocket engine stops.

Uh = 62.5cos80 = 10.853 m/s

Then there is the S1=Uh*t for the aceleration stage:

(linear distance traveled under acceleration = s=0*t+1/2*25*2.52 = 78.125m)

If there is a better way of calculating this distance let me know?

Uh = 78.125cos80 = 13.57 m S1 horizontal

Answer:

S1 = 13.57m*2.5s = 33.925m
S2 = 10.853 *(6.27+7.42)= 148.58m

Therefore total distance = 182.505m

 

[Doc Al]

S=ut+1/2at²

Good. A more complete version would be:

y = y₀ + ut + 1/2at²

You know:
- the initial y position: y₀
- the acceleration
- the final y position (when it hits the ground)
- the initial velocity (y component): u

That's all you need to solve for t, the time it takes to reach the ground.

 

[omoplata]

(linear distance traveled under acceleration = s=0*t+1/2*25*2.52 = 78.125m)

If there is a better way of calculating this distance let me know?

Uh = 78.125cos80 = 13.57 m S1 horizontal

Answer:

S1 = 13.57m*2.5s = 33.925m
S2 = 10.853 *(6.27+7.42)= 148.58m

Therefore total distance = 182.505m

I agree with calculated linear distance traveled under acceleration, 78.125 m.

So the horizontal component of the linear distance traveled under acceleration should be 78.125*cos(80) . This should be S1. You don't need to multiply it again by 2.5 s. You can catch this error easily by looking at the units. Since you know that you will get a distance, the units of your final answer should be m. But the units for your S1 = 13.57m*2.5s is m*s.

I agree with your calculated value of S2.

Now all you have to do to find the total horizontal distance traveled is to add S2 to the correct value of S1.

 

[lubo]

I agree with calculated linear distance traveled under acceleration, 78.125 m.

So the horizontal component of the linear distance traveled under acceleration should be 78.125*cos(80) . This should be S1. You don't need to multiply it again by 2.5 s. You can catch this error easily by looking at the units. Since you know that you will get a distance, the units of your final answer should be m. But the units for your S1 = 13.57m*2.5s is m*s.

I agree with your calculated value of S2.

Now all you have to do to find the total horizontal distance traveled is to add S2 to the correct value of S1.

S1 = 13.57m = 13.57m
S2 = 10.853 *(6.27+7.42)= 148.58m

Therefore total distance = 162.15M

 

[omoplata]

Yes, I think this answer is correct.