Solving Sinusoidal Equations with Theta

[mcastillo356]

Homework Statement

How much is the minimum angle at wich a football thrown by a striker has to come out if he wants to save a 1,9 m player barrier that are located at a distance of 11 m, if you hit the ball with a initial speed of 100 km/h?

Relevant Equations

$v_x=v_0\cos{\theta}=d/t$
$v_y=v_0\sin{\theta}-gt$

I can't solve $\sin{\theta}\cos{\theta}=0,14$
Thanks!

 

[Tom.G]

I can't solve $\sin{\theta}\cos{\theta}=0,14$
Thanks!

Neither can I, but half a dozen tries on my calculator yielded 8.131°, equivalently 81.869°, both good to 4 digits.

 

[anuttarasammyak]

How did you get the relation
$$\sin\theta \cos\theta = 0.14$$?　I got a quadratic equation of $tan\theta$.

 

[archaic]

Haven't verified the physics, but you can use this:
$$\sin x\cos x=\frac 12\sin2x$$
You can find these identities using:
$$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$
where $i^2=-1$.
For your case, I just multiplied both functions.

 

[haruspex]

How did you get the relation
$$\sin\theta \cos\theta = 0.14$$?　I got a quadratic equation of $tan\theta$.

$2\sin(\theta)\cos(\theta)=\sin(2\theta)=\frac{2t}{1+t^2}$, where $t=\tan(\theta)$.
Also, $\cos(2\theta)=\frac{1-t^2}{1+t^2}$ and so $\tan(2\theta)=\frac{2t}{1-t^2}$.
These formulae are worth remembering (I remember very few).

 

[mcastillo356]

Hi anuttarasammyak, Tom.G. archaic, the problem is that I can't manage with the scientific calculator. I already know the formulae you've given. Wich steps must I make to do it on my own?
It's this way
$v_x=v_0\cos{\theta}=d/t$
$v_y=v_0\sin{\theta}-gt$
On top of parabolic movement, $v_y=0$
$v_0\sin{\theta}=gt$, so $\sin{\theta}=gt/v_0$
Clearing $t$ from $v_x$ equation
$\sin{\theta}=\dfrac{g\cdot{d}}{v_0^2\cdot{\cos{\theta}}}$; so
$\sin{\theta}\cos{\theta}=\dfrac{g\cdot{d}}{v_0^2}$
Let's see if I've done the wright the LaTeX; I can't seem to preview
haruspex, the formulae you've given are fantastic, but, how can I manage to obtain $\theta$ with the scientific calculator?

 

[archaic]

Hi anuttarasammyak, Tom.G. archaic, the problem is that I can't manage with the scientific calculator. I already know the formulae you've given. Wich steps must I make to do it on my own?
It's this way
$v_x=v_0\cos{\theta}=d/t$
$v_y=v_0\sin{\theta}-gt$
On top of parabolic movement, $v_y=0$
$v_0\sin{\theta}=gt$, so $\sin{\theta}=gt/v_0$
Clearing $t$ from $v_x$ equation
$\sin{\theta}=\dfrac{g\cdot{d}}{v_0^2\cdot{\cos{\theta}}}$; so
$\sin{\theta}\cos{\theta}=\dfrac{g\cdot{d}}{v_0^2}$
Let's see if I've done the wright the LaTeX; I can't seem to preview
haruspex, the formulae you've given are fantastic, but, how can I manage to obtain $\theta$ with the scientific calculator?

$$\frac 12\sin2x=0.14\implies2x=\arcsin{0.28}\implies x=\frac12\arcsin{0.28}$$
This will give a value of $x$ in $[-\pi/2,\,\pi/2]$.
The arcsine function is available in scientific calculators.

 

[etotheipi]

The best way is without a doubt @archaic's. You could also try
$$\sin^2 \theta \cos^2 \theta = \sin^2 \theta (1-\sin^2 \theta) = 0.14^2$$ and let $u = \sin^2 \theta$ to get a quadratic in $u$.

 

[mcastillo356]

Hi, archaic, forum
My calculator gives 8,13010... degrees. Correct?

 

[anuttarasammyak]

It's this way

Where in your way "1,9 m player barrier" is considered ?

 

[mcastillo356]

Hi, anuttarasammyak
Not at all. Strange, isn't it?
It's all about an already solved problem by the spanish UNED. I'm preparing the access exam previous to get matriculated on the grade of maths

 

[anuttarasammyak]

I am afraid you have a wrong solution. For hobbit barriers and for Godzilla barriers you kick the ball at the same angle ? You had better ask it to your teacher.

 

[mcastillo356]

Haha...!. Yes indeed...!. I will try to solve it my way. I'm back today. Now it's 12.24 PM in Spain. Please, forum, give me time

 

[haruspex]

On top of parabolic movement,

You cannot assume that is where it must be at 1,9m.
Even if the question were to find the minimum launch velocity to clear the player that would not be a correct assumption.

 

[Delta2]

You cannot assume that is where it must be at 1,9m

It might be the case that the universe (and the data given) conspire and the peak of that parabola lies exactly 1.9m above the ground...

 

[ehild]

$2\sin(\theta)\cos(\theta)=\sin(2\theta)=\frac{2t}{1+t^2}$, where $t=\tan(\theta)$.
Also, $\cos(2\theta)=\frac{1-t^2}{1+t^2}$ and so $\tan(2\theta)=\frac{2t}{1-t^2}$.
These formulae are worth remembering (I remember very few).

Also it is an other important formula worth remembering:
$$\frac{1}{\cos^2(\theta)}=1+\tan^2(\theta)$$
With this one, you get a quadratic equation in theta.

 

[mcastillo356]

I've written to my teacher. Let's see what says.

 

[mcastillo356]

Thanks berkeman!.

 

[haruspex]

I've written to my teacher. Let's see what says.

Why? Let's just try to solve it correctly.
You know that after traveling x=11m horizontally it needs to be at height y=1,9m.
Suppose it takes time t to get there.
If you kick it at speed v and at angle theta above the horizontal, what will x and y be at time t?
What two equations does that allow you to write?

 

[mcastillo356]

Hi haruspex!
I've already done it by myself: it's $0,09\pi$rad

 

[haruspex]

Hi haruspex!
I've already done it by myself: it's $0,09\pi$rad

Seems too much. Please post your working.

 

[mcastillo356]

I think that things missing, I mean, the distance, for example, is at the statement of the problem; the height of the barrier is implicit in $\sin{\theta}$ and $\cos{\theta}$. So:

$v_x=v_0\cos{\theta}=d/t$
$v_y=v_0\sin{\theta}-gt$
On top of parabolic movement, $v_y=0$
$v_0\sin{\theta}=gt$, so $\sin{\theta}=gt/v_0$
Clearing $t$ from $v_x$ equation
$\sin{\theta}=\dfrac{g\cdot{d}}{v_0^2\cdot{\cos{\theta}}}$; so
$\sin{\theta}\cos{\theta}=\dfrac{g\cdot{d}}{v_0^2}$

$$\frac 12\sin2x=0.14\implies2x=\arcsin{0.28}\implies x=\frac12\arcsin{0.28}$$
This will give a value of $x$ in $[-\pi/2,\,\pi/2]$.
The arcsine function is available in scientific calculators.

And then:
$\arcsin{0,28}=16,26$ degrees;
Degrees to radians:
$\dfrac{16,26º\pi\mbox{rad}}{180º}=0,09\pi\mbox{rad}$

Pros and cons:
Am I talking to myself?; 16,26 are degrees?; what the hell is this diabolic statement of the problem?;
On the other hand: the result is more than $-2\pi$ and less than $2\pi$

Seems too much

Yes

 

[haruspex]

I think that things missing, I mean, the distance, for example, is at the statement of the problem; the height of the barrier is implicit in $\sin{\theta}$ and $\cos{\theta}$. So:

And then:
$\arcsin{0,28}=16,26$ degrees;
Degrees to radians:
$\dfrac{16,26º\pi\mbox{rad}}{180º}=0,09\pi\mbox{rad}$

Pros and cons:
Am I talking to myself?; 16,26 are degrees?; what the hell is this diabolic statement of the problem?;
On the other hand: the result is more than $-2\pi$ and less than $2\pi$
Yes

Your answer is wrong. As I pointed out in post #14, you cannot assume that 1,9m is the top of the arc.

Instead, follow the system I laid out in post #19.

 

[etotheipi]

A useful shortcut for projectile motion questions like this is the general form of the parabolic trajectory $$y = x\tan{\theta} - \frac{gx^2}{2u^2 \cos^2{\theta}}$$although I wouldn't use it unless you can derive it first. It comes from combining the two equations for $x(t)$ and $y(t)$ w/ uniform acceleration of magnitude $g$ in the $-\hat{y}$ direction.

 

[mcastillo356]

Here is the image of the parabolic movement. Excuse my spanglish ("parabolic movement" is the words for this image?)

 

[mcastillo356]

Ooops... Hope not to done something not allowed

 

[etotheipi]

Like @haruspex mentioned, you can't assume that the ball crosses the barrier at the highest point of its trajectory. You are expecting solutions that look something like this:

Generally you will have 2 distinct solutions for $\theta$, corresponding to each trajectory. Only if these both happen to be the same will highest point on the trajectory be at the barrier.

 

[mcastillo356]

Wow...!. It's true... It might travel to the Moon and save the barrier, and score

 

[mcastillo356]

Why? Let's just try to solve it correctly.
You know that after traveling x=11m horizontally it needs to be at height y=1,9m.
Suppose it takes time t to get there.
If you kick it at speed v and at angle theta above the horizontal, what will x and y be at time t?
What two equations does that allow you to write?

Maybe $v_f^2=v_i^2-2g\Delta{y}$
$\Delta{x}=v_o\cos{\theta}\cdot{t}$
No

 

[mcastillo356]

Tomorrow I will try once again. It's 02:08 AM.

 

[haruspex]

A useful shortcut for projectile motion questions like this is the general form of the parabolic trajectory $$y = x\tan{\theta} - \frac{gx^2}{2u^2 \cos^2{\theta}}$$although I wouldn't use it unless you can derive it first. It comes from combining the two equations for $x(t)$ and $y(t)$ w/ uniform acceleration of magnitude $g$ in the $-\hat{y}$ direction.

Since it is the angle that is to be found, it is a little more convenient in the form $$y = x\tan{\theta} - \frac{gx^2}{2u^2}(1+ \tan^2{\theta})$$

 

[haruspex]

Maybe $v_f^2=v_i^2-2g\Delta{y}$

That's not very useful because you do not know the vertical velocity as it passes over the defenders. Use the formula involving time instead.

$\Delta{x}=v_o\cos{\theta}\cdot{t}$

Yes.

 

[mcastillo356]

Hi, forum!

Why? Let's just try to solve it correctly.
You know that after traveling x=11m horizontally it needs to be at height y=1,9m.
Suppose it takes time t to get there.
If you kick it at speed v and at angle theta above the horizontal, what will x and y be at time t?
What two equations does that allow you to write?

Kinematic equations for projectiles:
$x=v_0\cos{\theta}\cdot{t}$
$y=v_0\sin{\theta}\cdot{t}-1/2g\cdot{t}^2$
Hi haruspex!, are these ones?

 

[Delta2]

Solve the first equation for time t and replace t in the second equation, what do you get?

 

[archaic]

Player: *kicks the ball*.
Let's assume that the initial velocity vector is such that when the ball is at the same horizontal position (from its starting point) as the player wall, it is just on top of it (this is the case where the angle is a minimum). Let's call this instant $\tau$.
When I do the math (which you did), I find that the vertical position of the ball at any given instant is described by $y(t)=(v_0\sin{\theta})t-\frac12gt^2$, hence $y(\tau)=1.9=(v_0\sin{\theta})\tau-\frac12g\tau^2$.
The problem statement wants $\theta$, so let's solve for it!
$$\begin{align*}
(v_0\sin{\theta})\tau-\frac12g\tau^2=1.9&\Leftrightarrow v_0\sin{\theta}-\frac12g\tau=\frac{1.9}{\tau}\text{, we know that $\tau\neq0$}\\
&\Leftrightarrow v_0\sin{\theta}=\frac{1.9}{\tau}+\frac12g\tau\\
&\Leftrightarrow \sin{\theta}=\frac{1.9}{v_0\tau}+\frac{1}{2v_0}g\tau\\
&\Leftrightarrow \theta=\arcsin\left(\frac{1.9}{v_0\tau}+\frac{1}{2v_0}g\tau\right)\\
\end{align*}$$
It seems that we're missing $\tau$, but we can find it!
The horizontal position is given by $x(t)=(v_0\cos{\theta})t$, thus, at $t=\tau$, we have $x(\tau)=11=(v_0\cos{\theta})\tau$, which gives us $\tau=\frac{11}{(v_0\cos{\theta})}$. Let's plug it in $\theta$:
$$\theta=\arcsin\left(\frac{1.9\cos\theta}{11}+\frac{11g}{2v_0^2\cos\theta}\right)$$
Scary . Let's take a step back!
$$\sin\theta=\frac{1.9\cos\theta}{11}+\frac{11g}{2v_0^2\cos\theta}$$
Frankly, at this point, you might not see what you should do if you haven't seen the spatial equation of a projectile and recognized the form.
If we divide by $\cos\theta$ (which is definitely not $0$), we obtain a $\tan\theta$ on the left, and a $1/\cos^2\theta$ on the right, which is equal to $1+\tan^2\theta$.
$$\tan\theta=\left(\frac{1.9}{11}+\frac{11g}{2v_0^2}\right)+\left(\frac{11g}{2v_0^2}\right)\tan^2\theta$$
This is @haruspex #31 post. You can, now, solve for $\tan\theta$, then find $\theta$.