Hitting a rocket with a projectile

[MichaelTam]

Homework Statement

Test

Relevant Equations

$a[t]= A - B t^2$

A person standing a distance 𝑑 from the rocket launch site shoots a projectile at 𝑡=0 at an initial speed 𝑣0 at an angle 𝜃0 with respect to the horizontal as shown in the figure above. The projectile hits the rocket just when the rocket reaches its maximum height. The downward gravitational acceleration is 𝑔.

(Part b) Find an expression for the tangent of the angle tan(𝜃0) in terms of 𝐴, 𝐵, 𝑑, and 𝑔. Do not use 𝑣0 in your answer.
,In part a, I use the formula for finding $t_f = \sqrt {3 A/B}$ where $t_f$ is the time the rocket approach to the maximum height where the velocity will become zero due to the gravity on the rocket, where now the question can’t let me use $v_0$ as one of the answer’s variable ,according to the Pythagorean’s Therom , $v_0^2=v_0,y^2+v_0,x^2$or $\tan \theta=\frac {v_0,y} {v_0,x}$, so my first goal is to find out the $v_0$ firs.
My assumption according to the question is the initial position of the rocket is $x=d, y = 0$
the initial position of the stone is $x=0 , y=0$
The velocity of the rocket approaches the highest point,$v t_f = A t_f - \frac {B t_f^3} {3} = 0$
However, I don’t have the information of the stone velocity when it approaches $t_f$
But the height of the stone and the rocket should be the same at $t_f$ so...
Velocity of the stone at time equals to $t_f$

$\frac {A t_f^2} {2} - \frac {B t_f^4} {4} = v_y,0 - g t_f^2/2$
after substitute the $t_f$equation and simplify ,I get $v_y,0= (\sqrt {48 A g ({g-A}) + 3 A })/4$
For $v_x,0$, $v_x,0=d/t_f$
substitute $t_f$ equation ,I get
$v_x,0= \frac {d \sqrt {3AB}} {3 A}$
using $\tan \theta=\frac {v_0,y} {v_0,x}$, and substitute the result and simplify,I get
$\tan \theta = \frac {\sqrt {144 A^2 g^2 - 144 A^3 + 9 A^2 B}} {4Bd}$
I don’t sure that if my step and assumption is correct, because I am not confident on two dimensions, especially the projectile because I have few confident on dealing with the algebra too...

 

[MichaelTam]

Can anyone help me to check my solution?I am not sure if my process is correct...

 

[haruspex]

Where does this equation come from?
Relevant Equations: $a[t]=A−Bt^2$
Is it provided with the question as the acceleration of the rocket?
If so, please include that part of the question, and do not list it as a Relevant Equation. That section is for standard physics equations.
And please do not duplicate threads. Just post a correction within the thread.

 

[MichaelTam]

Yes, that’s the acceleration of the rocket

 

[haruspex]

Yes, that’s the acceleration of the rocket

Is it purely vertical?

 

[MichaelTam]

Yes, no air resistance, purely vertical acceleration, no horizontal motion is describe on the rocket.

 

[haruspex]

I do not understand this step. Looks like you intended to integrate the velocity to find the height.

The velocity of the rocket approaches the highest point,$v t_f = A t_f - \frac {B t_f^3} {3} = 0$
:
$\frac {A t_f^2} {2} - \frac {B t_f^4} {4} = v_y,0 - g t_f^2/2$

 

[MichaelTam]

Yes, I do the indefinite integral in order to get the velocity and the position , because the constant is zero, so I didn’t add the constant.

 

[haruspex]

Yes, I do the indefinite integral in order to get the velocity and the position , because the constant is zero, so I didn’t add the constant.

But the second integration has an error, and you wrote it equal to v_y instead of to a height.

 

[MichaelTam]

What is the error...,you mean the integration of velocity?If it is, I had follow the rules of integration, $\int x^ n dx$ = $\frac {x^n+1} {n+1}$

 

[haruspex]

What is the error...,you mean the integration of velocity?

Integrating the velocity $A t_f - \frac {B t_f^3} {3}$ does not produce $\frac {A t_f^2} {2} - \frac {B t_f^4} {4}$, and even if integrated correctly it would not produce a velocity, but you wrote $\frac {A t_f^2} {2} - \frac {B t_f^4} {4} = v_y$.

 

[MichaelTam]

Ops, sorry i just find the miss typing variable, the $v_0,y$should be also multiply by the $t_f$.

 

[haruspex]

What is the error...,you mean the integration of velocity?If it is, I had follow the rules of integration, $\int x^ n dx$ = $\frac {x^n+1} {n+1}$

No, you have not applied it correctly. What happened to the $\frac 13$?
If you cannot spot your error, differentiate back to see if you get the original.

 

[haruspex]

Ops, sorry i just find the miss typing variable, the $v_0,y$should be also multiply by the $t_f$.

Ah, you meant $\frac {A t_f^2} {2} - \frac {B t_f^4} {4} = v_{y,0}t_f - g t_f^2/2$.
You need to put { and } around the subscripts on the v, otherwise only the y becomes a subscript.

 

[MichaelTam]

I mean the $\iint a_r(t) = v_y,0 t_f - (g t_f^2/2)$ $a_r$ is the acceleration of the rocket.
Because integrate the stone vertical velocity will get its position respect to time $t_f$ on the left side, the meaning of the equation is the right hand side of the position of the stone will as same as the left side of the position of the rocket due to the time of impact.

 

[MichaelTam]

Sorry, but what is {and} means...you mean this is a function to put the terms together?

 

[haruspex]

Sorry, but what is {and} means...you mean this is a function to put the terms together?

E.g. in your post #15 you have $\iint a_r(t) = v_y,0 t_f - (g t_f^2/2)$. The $v_y,0$ makes no sense, and threw me off. You meant $v_{y,0}$.
To see the LaTeX difference, hit reply on this post.

Have you spotted your integration error yet?

 

[MichaelTam]

Ok, I make the correct integration now!
$\frac {At^2} {2} - \frac {Bt^4} {12} = v_{y,0}t_f - \frac {gt^2} 2$ where the $v_y$is also the initial velocity $v_0$, so what is the next check?

 

[haruspex]

Ok, I make the correct integration now!
$\frac {At^2} {2} - \frac {Bt^4} {12} = v_{y,0}t_f - \frac {gt^2} 2$ where the $v_y$is also the initial velocity $v_0$, so what is the next check?

Don't you mean $v_{y,0}=v_0\sin(\theta_0)$?
What answer do you get now?

 

[MichaelTam]

Still calculating...yes... and also $v_{x,0} = v_0\cos (\theta_0)$
$v_{y,0}= \frac{\sqrt {\frac {A( 3 A + g )^2} {3B}}} 4$
so, because the error does not affect $v_{x,0}$
so it is still $= \frac {d \sqrt{3AB}} {3A}$
plug the result back again into $\tan (\theta_0)= \frac { v_{y,0}} {v_{x,0}}$
I will get...

 

[MichaelTam]

Do I need to rationalise the answer?

 

[MichaelTam]

I get the final answer,

$\tan (\theta_0) = \frac {A(3A+g)} {4Bd}$

 

[MichaelTam]

So, how to check my answer correctly?

 

[MichaelTam]

Do I need to simplify it or just keep the answer unexpend.

 

[haruspex]

$v_{y,0}= \frac{\sqrt {\frac {A( 3 A + g )^2} {3B}}} 4$

That's not what I get. Please post your steps.

 

[MichaelTam]

Errrrrr...
$v_{y,0}= \frac {(3A+g)(\sqrt{3AB})} {12B}$

 

[MichaelTam]

 

[MichaelTam]

Is it correct now(the calculation)?

 

[MichaelTam]

So, my assumption is correct, the calculation is correct, what i can put the answer into which i can prove my answer is correct? Because i use this problem to get pass of my course...

 

[haruspex]

View attachment 269315

Mistake in line 5. You forgot to multiply the 3Ag/B term by 6 when expanding the parentheses.