Physics Equation - cannot solve for $\theta$

Question

I'm trying to solve the first part of this problem and my trigonometry skill aren't as honed in as they used to be, could anyone help me out?

While standing on friction less ice, you($m = 65.0 \textrm{ kg}$) toss a $4.50 \textrm{ kg}$ rock with initial speed $12 \textrm{ m/s}$. The rock is $15.2 \textrm{ m}$ from you when it lands. Take the $x$-axis on the ice surface in the horizontal direction of the motion of the rock. Ignore the initial height of the toss. AT WHAT ANGLE DID YOU TOSS IT?

You can solve this by using your givens and solving for $\theta$ in the angular trajectory equation:

$$x = \frac{v^2}{g}\sin(2\theta)$$

Please don't give me the answer, I just want to see if anyone can solve this equation for $\theta$.

Thank you!

Answer 1

In order to solve this you need to look at inverse trigonometric functions:

To solve for $\theta$ in general:

$ \sin(\theta) = y \hspace{5 mm} \Rightarrow \hspace{5 mm} \theta=\sin^{-1}(y) $

For your example:

$ \frac{gx}{v^2} = \sin(2\theta) $

Thus:

$2\theta = \sin^{-1}\left(\frac{gx}{v^2}\right)$

So:

$ \theta = \frac{1}{2} \sin^{-1}\left(\frac{gx}{v^2}\right) $

P.S. You can read more on inverse trig functions here.

Answer 2

You mean this solution:

$$ x = \frac{v^2}{g} \sin ( 2 \theta) \rightarrow \sin(2 \theta) = \frac{x\,g}{v^2} \rightarrow 2 \theta = \sin^{-1} \left( \frac{x\,g}{v^2} \right) $$

$$\boxed{ \theta = \frac{1}{2} \sin^{-1} \left( \frac{x \,g}{v^2} \right)} $$

Answer 3

You must use conservation of momentum to determine the distance that the rock actually travels from the point it was thrown (this will be different than 15.2m). Once you have this, you can use the inverse sine function to determine the angle.

Answer 4

Regarding your comment on getting started off, I think this should help clarify and better motivate the formula you are given, in the $x$-direction, we can use conservation of linear momentum. Before the throw, neither mass is moving, so we have $p_{initial}=0$. Let $M$ be your mass and $m$ the mass of the ball, after the throw, we have $p_{final} = Mv + mv_0 \cos(\theta)$, where $v$ is your velocity in the $x$ direction and $v_0 =12 \: m/s$ we can then write: $$ p_{initial} = p_{final}$$ $$ 0 = Mv + mv_0 \cos(\theta) $$ $$\implies v = - \frac{m v_0 \cos(\theta)}{M}$$ Your displacement is given by: $$ d_M = vt$$ and the displacement of the ball in the $x$ direction is given by: $$d_m = v_0 t$$ You are given $\Delta d = |d_m - d_M| = d_m -d_M =15.2\: m$, now, in the $y$ direction, the ball will move in a parabolic arc with final displacement equal to $0$, thus: $$ 0 = v_0 \sin(\theta) t - \frac{1}{2} g t^2$$ Solving for $t$, we have: $$ t = \frac{-v_0 \sin(\theta) \pm \sqrt{v_0^2 \sin^2(\theta)}}{-g}$$ Where we see that the only non-zero solution is: $$ t = \frac{2v_0 \sin(\theta)}{g}$$ If you plug this into your formula for $\Delta d$, you have: $$ \Delta d = v_0\cos(\theta) t - v t = v_0 \cos(\theta) \left(\frac{2v_0 \sin(\theta)}{g} \right) - v\left(\frac{2v_0 \sin(\theta)}{g} \right) $$ $$ = v_0 \cos(\theta) \left(\frac{2v_0 \sin(\theta)}{g} \right) + \frac{m v_0 \cos(\theta)}{M}\left(\frac{2v_0 \sin(\theta)}{g} \right) $$

$$ =2\cos(\theta) \sin(\theta) \left(\frac{v_0^2}{g} +\frac{m v_0^2 }{Mg} \right) $$ You can then use the trig identity (which you should try to derive, to strengthen your trig skills) $2\sin(\phi)\cos(\phi) = \sin(2\phi)$ Which gives the formula: $$ \Delta d = \sin(2 \theta ) \left(-\frac{m v_0^2 }{Mg} + \frac{v_0^2}{g}\right)$$ Where you can now use the inverse sine function mentioned in other replies to find the angle that the ball is launched, $\theta$. Note that there is a correction term to the formula you were given due to the movement of the person throwing the ball. We can also check limiting cases, which is always a good idea in physics when confirming your answer is reasonable. If your mass $M \to \infty$ then the first term: $$ \frac{m v_0^2 }{Mg} \to 0$$ So you recover the usual formula: $$ \sin(2 \theta ) \left( \frac{v_0^2}{g}\right)$$ If $M \to 0$, you would expect that by conservation of momentum, you would slide really far away, which we see is the case in our formula.