- #36
mcastillo356
Gold Member
- 560
- 267
Two solutions:archaic said: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##.
##\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