Why is my calculation for a rocket's maximum range incorrect?

[yesmale4]

Homework Statement

A 1500 kg rocket is launched with a velocity v0 = 115 m/s against a strong wind. The wind exerts a constant horizontal force Fwind = 1450 N on the rocket. What is the rockets maximum range?

Relevant Equations

R=(V0^2*sin(2a))/g

the first question was At what launch angle will the rocket achieve its maximum range?
and i found the right answer it was 42.186deg.
but now i don't understand what I am doing wrong i put v0,a,g in the equation of range and its wrong and i don't undertand why
here is what i try:
R=(115^2*sin(2*42.186))/9.81
and my answer is 1341.615m
and it was wrong so than i thought maybe i need to divide by 2 and it was still wrong i would like for some help with it because i don't know what else to do.

 

[TSny]

Relevant Equations:: R=(V0^2*sin(2a))/g

This is the formula for the range if there is no wind. You need to rederive the range when you do have wind.

 

[yesmale4]

This is the formula for the range if there is no wind. You need to rederive the range when you do have wind.

thank you for the answer but i still don't understand, I searched for the formula of the derivative of the range and all I got was the formula I wrote down here for example
so i don't understand which formula do you talking about

 

[TSny]

The formula for the horizontal range derived in the video is based on the assumption that there is no wind. So, it is not applicable to your problem where there is a wind that produces a constant horizontal force on the rocket.

When there is no wind, the horizontal speed of the projectile remains constant. That is, there is no horizontal acceleration. That is why the video bases its derivation of the horizontal range on the following simple equation:

horizontal range = horizontal speed x time

or

$R = (v_0 \cos \theta) \, t$, $\;\;$ where $\theta$ is the launch angle, $v_0$ is the launch speed, and $t$ is the time of flight.

But in your problem, you have a horizontal wind that produces a constant horizontal force on the rocket. Can you calculate the horizontal acceleration $a_x$ produced by this force? You will need to think about the sign of $a_x$ as the problem states that the rocket is launched "against" the wind.

Can you see how the equation $R = (v_0 \cos \theta) \, t$ should be modified in order to include the acceleration $a_x$?

Edit: I'm also curious as to how you were able to work out the answer for the angle of projection that produces the maximum range. Can you summarize what you did for this part?

 

[yesmale4]

The formula for the horizontal range derived in the video is based on the assumption that there is no wind. So, it is not applicable to your problem where there is a wind that produces a constant horizontal force on the rocket.

When there is no wind, the horizontal speed of the projectile remains constant. That is, there is no horizontal acceleration. That is why the video bases its derivation of the horizontal range on the following simple equation:

horizontal range = horizontal speed x time

or

$R = (v_0 \cos \theta) \, t$, $\;\;$ where $\theta$ is the launch angle, $v_0$ is the launch speed, and $t$ is the time of flight.

But in your problem, you have a horizontal wind that produces a constant horizontal force on the rocket. Can you calculate the horizontal acceleration $a_x$ produced by this force? You will need to think about the sign of $a_x$ as the problem states that the rocket is launched "against" the wind.

Can you see how the equation $R = (v_0 \cos \theta) \, t$ should be modified in order to include the acceleration $a_x$?

Edit: I'm also curious as to how you were able to work out the answer for the angle of projection that produces the maximum range. Can you summarize what you did for this part?

yes of course about how i found the angle:
i did x0 + v0t+(1/2g*t^2)*(Fw/mg)
i found an equation than i did derivative because they ask about maxium
after i did that i compared the equation to 0 and found x which was 42.186.

about what you said i know how to find the acceleration, about your equation i believes i need to put another equation insted of t but i can't think of any equation that fit, the only one who come to my mind is t=(v-v0)/a but i don't know what is v .
i am really hope that you can help me with that.

 

[haruspex]

yes of course about how i found the angle:
i did x0 + v0t+(1/2g*t^2)*(Fw/mg)
i found an equation than i did derivative because they ask about maxium
after i did that i compared the equation to 0 and found x which was 42.186.

about what you said i know how to find the acceleration, about your equation i believes i need to put another equation insted of t but i can't think of any equation that fit, the only one who come to my mind is t=(v-v0)/a but i don't know what is v .
i am really hope that you can help me with that.

That equation, assuming Fw is being given a negative value because the rocket is launched into the wind, gives the x displacement at time t.
You haven't explained how you then involved the launch angle. Please post the rest of your working so that we can advise on the how to find the range from what you have already done.

 

[yesmale4]

That equation, assuming Fw is being given a negative value because the rocket is launched into the wind, gives the x displacement at time t.
You haven't explained how you then involved the launch angle. Please post the rest of your working so that we can advise on the how to find the range from what you have already done.

i hope its ok.
i try to put the t i found in TSny equation -> R=(V0*cos(x))*t but again i got the same result of 1341.615m

 

[TSny]

Your work shown in the handwritten part of your previous post looks to me like it is correct. But, honestly, it's a notational nightmare.

Your expression for the time of flight $t = \frac{2v_0\sin(x)}{g}$ is correct, although you don't derive it in your work. Here, you are using $x$ for the unknown launch angle.

Then you write an equation for the horizontal displacement traveled and you use $x$ for both the horizontal distance and the unknown angle:

It looks correct except I don't know what the symbol that I circled in green represents.

Then it appears that you switch from using $x$ as the horizontal displacement to using $y$ as the horizontal displacement. (Often $y$ is used for the vertical displacement.) You then use calculus to find the angle that yields the maximum range. It looks correct.

i try to put the t i found in TSny equation -> R=(V0*cos(x))*t but again i got the same result of 1341.615m

As I've pointed out a couple of times before, this equation for $R$ does not apply to this problem because this equation does not take into account the wind. But you already have derived the correct equation for the horizontal displacement:

where y is denoting horizontal displacement.

Can you use this to get the maximum horizontal range, R?

 

[yesmale4]

Your work shown in the handwritten part of your previous post looks to me like it is correct. But, honestly, it's a notational nightmare.

Your expression for the time of flight $t = \frac{2v_0\sin(x)}{g}$ is correct, although you don't derive it in your work. Here, you are using $x$ for the unknown launch angle.

Then you write an equation for the horizontal displacement traveled and you use $x$ for both the horizontal distance and the unknown angle:

View attachment 298612

It looks correct except I don't know what the symbol that I circled in green represents.

Then it appears that you switch from using $x$ as the horizontal displacement to using $y$ as the horizontal displacement. (Often $y$ is used for the vertical displacement.) You then use calculus to find the angle that yields the maximum range. It looks correct.As I've pointed out a couple of times before, this equation for $R$ does not apply to this problem because this equation does not take into account the wind. But you already have derived the correct equation for the horizontal displacement:
View attachment 298613 where y is denoting horizontal displacement.

Can you use this to get the maximum horizontal range, R?

No unfortunately i don't know how to find the range from the horizontal displacement equation

 

[TSny]

No unfortunately i don't know how to find the range from the horizontal displacement equation

Can you define what the word "range" means in this context?

 

[yesmale4]

Can you define what the word "range" means in this context?

Ohh the horizontal distance i understand so i just need to find the X of this equation?
becuse in order to find the angle i derivative and then find the X so it the same but this time i don't need to do derivative?

 

[TSny]

Ohh the horizontal distance i understand so i just need to find the X of this equation?
becuse in order to find the angle i derivative and then find the X so it the same but this time i don't need to do derivative?

Right. You have found the angle that yields the maximum X (horizontal distance). Use this angle to calculate X.

 

[yesmale4]

Right. You have found the angle that yields the maximum X (horizontal distance). Use this angle to calculate X.

i don't know why but it still say i got wrong answer i did:
1348.144*sin(2*42.186)-265.683sin^2(42.186)
and got 467.78 and its the wrong answer, what i do wrong?

 

[TSny]

i don't know why but it still say i got wrong answer i did:
1348.144*sin(2*42.186)-265.683sin^2(42.186)
and got 467.78 and its the wrong answer, what i do wrong?

When I evaluate 1348.144*sin(2*42.186^o)-265.683sin^2(42.186^o) I don't get 467.78.

Check your evaluation of this expression.

 

[haruspex]

i don't know why but it still say i got wrong answer i did:
1348.144*sin(2*42.186)-265.683sin^2(42.186)
and got 467.78 and its the wrong answer, what i do wrong?

A good habit to get into is to work purely algebraically, only plugging numbers at the end. It has many advantages, including reducing the risk of numerical errors.
In the present case, you could have arrived at the range being $(\frac vg)^2\sqrt {g^2+a^2}$, where $F_w=ma$.

 

[TSny]

In the present case, you could have arrived at the range being $(\frac vg)^2\sqrt {g^2+a^2}$, where $F_w=ma$.

I don't quite get this expression for the maximum upwind range. I get a slightly more complicated expression. I would expect the max upwind range for a very strong wind ($a \gg g$) to be small.

 

[haruspex]

I don't quite get this expression for the maximum upwind range. I get a slightly more complicated expression. I would expect the max upwind range for a very strong wind ($a \gg g$) to be small.

Thanks for catching my typo. It was supposed to be $(\frac vg)^2(\sqrt {g^2+a^2}-a)$

 

[Lnewqban]

Thanks for catching my typo. It was supposed to be (vg)2(g2+a2−a)

Would you mind to explain how did you arrive to that arrangement of both accelerations?
Also, isn't 45 degrees always the angle of maximum range?

 

[haruspex]

Would you mind to explain how did you arrive to that arrangement of both accelerations?
Also, isn't 45 degrees always the angle of maximum range?

The constant lateral force effectively changes the angle of gravity and increases its magnitude. This makes the question equivalent to "range on an inclined plane."
https://iitutor.com/the-trajectory-...-inclined-plane-with-maximum-range-explained/

 

[Lnewqban]

The constant lateral force effectively changes the angle of gravity and increases its magnitude. This makes the question equivalent to "range on an inclined plane."
https://iitutor.com/the-trajectory-...-inclined-plane-with-maximum-range-explained/

Thank you.
Based on that idea, could we rotate the axes of coordinates and the landing plane 5.62°, which is the angle formed between the net acceleration vector and the vertical?

 

[haruspex]

Thank you.
Based on that idea, could we rotate the axes of coordinates and the landing plane 5.62°, which is the angle formed between the net acceleration vector and the vertical?

Yes, that’s what turns it into the "range on an inclined plane" version. The "range" is now a distance up the plane, not horizontal.
It is a curious fact that at maximum range the direction of launch is normal to the direction of landing, regardless of the plane's angle. I feel there should be an easy way to show that, but even if there is I don't see that it gives a quick way to find the formula.

 

[Lnewqban]

Yes, that’s what turns it into the "range on an inclined plane" version. The "range" is now a distance up the plane, not horizontal.
It is a curious fact that at maximum range the direction of launch is normal to the direction of landing, regardless of the plane's angle. I feel there should be an easy way to show that, but even if there is I don't see that it gives a quick way to find the formula.

Very interesting, haruspex.
Thank you very much.