Solving Sinusoidal Equations with Theta

[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 $\theta$.

 

[mcastillo356]

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 $\theta$.

Two solutions:
$\tan{\theta}=14$, so $\arctan{14}=85,91$;
$\tan{\theta}=0,29$, so $\arctan{0,29}=16,17$
I don't dare to conclude anything. Do you, forum?
haruspex, Delta2, I've chosen to explore haruspex and archaic's suggest

 

[haruspex]

Two solutions:
$\tan{\theta}=14$, so $\arctan{14}=85,91$;
$\tan{\theta}=0,29$, so $\arctan{0,29}=16,17$
I don't dare to conclude anything. Do you, forum?
haruspex, Delta2, I've chosen to explore haruspex and archaic's suggest

I got about 0,25 for the lower value of tan(θ). Did you plug your value back into the quadratic to check?

It is correct that there are two angles which will cause the ball to just clear the defenders. Any angle between the two will pass higher over their heads. The question asks for the minimum.

 

[archaic]

Two solutions:
$\tan{\theta}=14$, so $\arctan{14}=85,91$;
$\tan{\theta}=0,29$, so $\arctan{0,29}=16,17$
I don't dare to conclude anything. Do you, forum?
haruspex, Delta2, I've chosen to explore haruspex and archaic's suggest

I get $14.054$ and $0.247$ for $\tan\theta$ with $g=9.81$.

 

[mcastillo356]

Sorry, haruspex, forum, I've worked all this time with a value of $g$ not accurate: 9,88. Thanks, archaic, your latest message has make me notice it.
Salutes to everybody!

 

[archaic]

Sorry, haruspex, forum, I've worked all this time with a value of $g$ not accurate: 9,88. Thanks, archaic, your latest message has make me notice it.
Salutes to everybody!

you're welcome!