HELP with question plss

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Help me with this question please:

You need to post how far you have got - we can only give hints.

Im not asking for the answer, onlu the question. I dont know what its asking.

Im not asking for the answer, onlu the question. I dont know what its asking.

It's asking you to effectively model the projectile motion in Cartesian coordinates, derive the equation of the parabolic path where $y$ is the dependent variable, $x$ is the independent variable, and $V,\alpha$ are parameters. Then convert x,y into polar coordinates and rearrange into the form they give.

What does it mean by trajectory? Do you just take the horizontal/vertical components of V and use energy equations?

also what is r and theta for?

If you draw a diagram of this scenario, you can deduce that the horizontal component of the displacement is $x=Vt\cos \alpha$, and the vertical displacement is $y = Vt\sin \alpha - \dfrac{g}{2}t^2$.

You can rearrange the x displacement for $t = \dfrac{x}{V\cos \alpha}$ and substitute this into the y displacement equation.

You should get a result which descibes $y$ in terms of $x$ alongside the parameters $V,\alpha$. This is the equation of the trajectory of the particle.


Then you should recall polar coordinates. You can express every coordinate $(x,y)$ in the Cartesian plane via two different pieces of information; the distance to the point (x,y), which is $r$, and the angle that this distance makes with the horizontal, which is $\theta$.
Namely;

$x=r\cos \theta$
$y=r\sin \theta$

and you substitute these into the result for the trajectory, then rearrange for their result.

What does the (g/2)t² represent?

If you draw a diagram of this scenario, you can deduce that the horizontal component of the displacement is $x=Vt\cos \alpha$, and the vertical displacement is $y = Vt\sin \alpha - \dfrac{g}{2}t^2$.

You can rearrange the x displacement for $t = \dfrac{x}{V\cos \alpha}$ and substitute this into the y displacement equation.

You should get a result which descibes $y$ in terms of $x$ alongside the parameters $V,\alpha$. This is the equation of the trajectory of the particle.


Then you should recall polar coordinates. You can express every coordinate $(x,y)$ in the Cartesian plane via two different pieces of information; the distance to the point (x,y), which is $r$, and the angle that this distance makes with the horizontal, which is $\theta$.
Namely;

$x=r\cos \theta$
$y=r\sin \theta$

and you substitute these into the result for the trajectory, then rearrange for their result.

If you draw a diagram of this scenario, you can deduce that the horizontal component of the displacement is $x=Vt\cos \alpha$, and the vertical displacement is $y = Vt\sin \alpha - \dfrac{g}{2}t^2$.

You can rearrange the x displacement for $t = \dfrac{x}{V\cos \alpha}$ and substitute this into the y displacement equation.

You should get a result which descibes $y$ in terms of $x$ alongside the parameters $V,\alpha$. This is the equation of the trajectory of the particle.


Then you should recall polar coordinates. You can express every coordinate $(x,y)$ in the Cartesian plane via two different pieces of information; the distance to the point (x,y), which is $r$, and the angle that this distance makes with the horizontal, which is $\theta$.
Namely;

$x=r\cos \theta$
$y=r\sin \theta$

and you substitute these into the result for the trajectory, then rearrange for their result.

If you draw a diagram of this scenario, you can deduce that the horizontal component of the displacement is $x=Vt\cos \alpha$, and the vertical displacement is $y = Vt\sin \alpha - \dfrac{g}{2}t^2$.

You can rearrange the x displacement for $t = \dfrac{x}{V\cos \alpha}$ and substitute this into the y displacement equation.

You should get a result which descibes $y$ in terms of $x$ alongside the parameters $V,\alpha$. This is the equation of the trajectory of the particle.


Then you should recall polar coordinates. You can express every coordinate $(x,y)$ in the Cartesian plane via two different pieces of information; the distance to the point (x,y), which is $r$, and the angle that this distance makes with the horizontal, which is $\theta$.
Namely;

$x=r\cos \theta$
$y=r\sin \theta$

and you substitute these into the result for the trajectory, then rearrange for their result.

For the second part, would you differentiate r with respect to alpha or theta?

Range is described by $r>0$ when $\theta = 0$ ... so hopefully you can deduce what's happening and what to do.

I got an expression for dr/da and subbed in a=45° and then got left with sin(theta)

If theta = 0 then the dr/da = 0 and hence it is a maximum. Is that enough to say?

(Also why is theta=0 ?? )

Range is a distance measured along the horizontal axis from the point of projection.

Every point along this axis has $\theta = 0$ in polar coordinates.

So you set $\theta=0$, then solve $\dfrac{dr}{d\alpha} = 0$ to obtain $\alpha = 45$.

To be more rigorous, you would show that $\dfrac{d^2r}{d\alpha^2} < 0$ at $\alpha = 45$ hence it's a maximum.