Projectile motion, throwing a ball at a coconut

[Kot]

Homework Statement

You throw a ball at a coconut on a stand 4.0 m in front of you and 1.0 m above the point at which you release the ball. If you throw the ball at 10.0 m/s, at what angle(s) should you throw it? How long is the ball in the air in each case?

Homework Equations

$v=v_o + at \\ x=x_o + v_o t + \dfrac{1}{2}at^2$

The Attempt at a Solution

I am given the horizontal distance (4.0m), vertical distance (1.0m), and the horizontal initial velocity (10.0m/s). I am looking for the angle θ. I have looked over some videos online and I know that there should be two angles where I can hit the coconut (they are complementary angles). I have trouble finding the angle θ because there are two unknowns in my equation (t and θ). I have split the problem into x and y components.

$v_x = v_{ox} \\ x = x_o + v_o t \\ v_y = v_{oy} + a_y t \\ y = y_o + y_o t + \dfrac{1}{2}a_y t^2$

I also used $v_{ox} = v_o cosθ \\ v_{oy} = v_o sinθ$ and substituted into the original equations. I have tried to cancel out t and solve for θ but I end up with a equation that involves the tangent and cosine function which I don't think is correct. Is this the correct approach to solving this problem?

 

[Simon Bridge]

You havn't been very careful with your notation though:

$v_0x=v_0\cos\theta \Rightarrow x=\cos\theta$
... see what I mean. That can lead to other mistakes later on.
I see three in your kinematic equations ferinstance -so I cannot tell what you've done.
Presumably you mean $v_0x$ to be $v_{0x}$ or something?

Lets focus on solving for components, and get rid of as many subscripts as we can JIC.

Putting the initial velocity as $\small \vec{u}=u_x\vec{\imath}+u_y\vec{\jmath}$ as the thing you need to find, and putting $\small T$ for the time of flight, I can rewrite your equations:

$v_y = u_y-gT\\ v_x = u_x\\ y_f=y_i+ u_iT-\frac{1}{2}gT^2\\ x_f=x_i+u_xT$

Show your working. Start by listing what you know and what you don't know.

 

[Kot]

I mistyped the subscripts, it was supposed to be $v_{ox}$. I have found a way to relate tanθ and cos^2θ, which is $\dfrac{1}{1+tan^2 θ}$, I then substituted t for tanθ and found that it was a quadratic equation. I applied the quadratic formula then took the inverse tangent of those two values and got θ=26.28 and θ=77.76.

I know that if the initial height is the same as the final height, the angles should be complementary. Is this no longer the case since the coconut is 1.0m above the initial height?

 

[Simon Bridge]

I don't know why you need to relate tan with cos^2.
If you haven't used the difference in height for your calculation, then it is unlikely you've done it right.
What you have done does not appear to have anything to do my reply.
What's wrong with following the advise there?

 

[Kot]

I got my answer right before you posted your response sorry. From your equations

$v_y = u_y-gT\\ v_x = u_x\\ y_f=y_i+ u_iT-\frac{1}{2}gT^2\\ x_f=x_i+u_xT$

I used the fact that $v_{ox} = 10.0m/s cosθ \\ v_{oy} = 10.om/s sinθ$. I then substituted $v_{ox}$ and $v_{oy}$ into $u_x$ and $u_i$ respectively.

$x_f=x_i+u_xT \\ 4.0m/s = 10.0m/s cosθ T$

and

$y_f=y_i+ u_iT-\frac{1}{2}gT^2 \\ 1.0m/s = 10.0m/s sinθ t - \frac{1}{2}(-9.8m/s^2)T^2$

I then solved for T in $4.0m/s = 10.0m/s cosθ T \\ T = \frac{2}{5cosθ}$ and plugged this into T in this equation $1.0m/s = 10.0m/s sinθ t - \frac{1}{2}(-9.8m/s^2)T^2$

Doing that, I ended up with $1.0m/s = 4tanθ - \frac{-9.8m/s^2}{(5cosθ)^2} \\ 1.0m/s = 4tanθ - \frac{-9.8m/s^2}{25cos^2θ}$

This is why I had to relate cos^2θ with tanθ. I substituted t for tanθ (t does not equal time, just a dummy variable) and found that it was a quadratic. Applying the quadratic formula I got my two angles mentioned previously.

 

[haruspex]

##
$1.0m/s = 4tanθ - \frac{-9.8m/s^2}{25cos^2θ}$

The units on the LHS should be m, not m/s.
I think you've dropped a factor of 2 in substituting for T in the T² term.

 

[Simon Bridge]

I got my answer right before you posted your response sorry. From your equations

$v_y = u_y-gT\\ v_x = u_x\\ y_f=y_i+ u_iT-\frac{1}{2}gT^2\\ x_f=x_i+u_xT$

I used the fact that $$v_{ox} = 10.0m/s cosθ \\ v_{oy} = 10.om/s sinθ$$.

I then substituted $v_{ox}$ and $v_{oy}$ into $u_x$ and $u_i$ respectively.

$v_{0x}=v_0\cos\theta \; : v_{0}=|\vec{v}_0|=10\text{m/s}$ you mean?
Glad to see that the notation thing is just the typing.
It can really mess up your marks if you wrote like that in an assignment.

It's quite hard to tell what you did, i.e. hard to tell if you dropped that half or not. Presumably you didn't in your actual assignment you handed in. Cudos for using LaTeX. Considering you made such an effort - some pointers...

...algebra:

* it is best practice to do all your algebra before putting numbers in - exception: trig functions of nice angles - leave the surds in though. It makes it easier to troubleshoot and also easier for people to award high marks in long answers.

* try to end expressions with the trig functions, so Tcosθ is fine, but you saw there was trouble with cosθT ... is the T inside or outside the cosine function? If you must, make what you mean clear like this: cos(θ)T ... rather than make people guess. Sometimes its not just one variable in there i.e. cos(t/(2πT)) - you can see how confusing it can get.

* you need to make it clear when a line of math leads from the previous line as opposed to just being another equation - like in simultanious equations. You can do this with \Rightarrow $\Rightarrow$ when one leads to the other.

* numbering equations means you don't have to write the whole thing out again - you can just say eq(5) or whatever.

... since LaTeX is so so useful:

* special functions in LaTeX are formatted properly (compare my cosine to yours) by using the backslash form. \cos is cosine, cos is three variables multiplied together.

* greek letters are a backslash followed by the name of the letter, thus:
theta is \theta or \Theta for $\theta$ or $\Theta$ ... see?
so cosθ is \cos\theta for $\cos\theta$

* units should be formatted as \text{} to avoid being confused with variables.
So - 10m/s is 10\text{ m/s} or 10\text{ms}^{-1} for $10\text{ m/s}$ or $10\text{ms}^{-1}$ depending which style you like to use.

* check the logic of your sentences - the formatting should make the meaning clearer.
You needed some extra newlines to make your meaning clear.

... overall:

* don't make the marker work so hard to understand you. The easier they understand you the more likely they will give you the benefit of the doubt and award you that extra mark.

Continuing your message: compare with what you wrote.

$x_f=x_i+u_xT \\ 4.0(\text{m/s}) = 10.0(\text{m/s}) T\cos\theta$

and

$y_f=y_i+ u_iT-\frac{1}{2}gT^2 \\ 1.0(\text{m/s}) = 10.0(\text{m/s}) t\sin\theta - \frac{1}{2}(-9.8)(\text{m/s}^2)T^2$

I then solved for T in $$4.0\text{m/s} = 10.0(\text{m/s}) T\cos\theta \\ T = \frac{2}{5\cos\theta}$$ ...and plugged this into T in this equation $$1.0(\text{m/s}) = 10.0 (\text{m/s}) t\sin\theta - \frac{1}{2}(-9.8)(\text{m/s}^2)T^2$$

Doing that, I ended up with $$1.0(\text{m/s}) = 4\tan\theta - \frac{-9.8(\text{m/s}^2)}{(5\cos\theta)^2} \\
\Leftrightarrow 1.0(\text{m/s}) = 4\tan\theta + \frac{9.8(\text{m/s}^2)}{25\cos^2\theta}$$

This is why I had to relate $\small \cos^2\theta$ with $\small \tan\theta$. I substituted t for tanθ (t does not equal time, just a dummy variable) and found that it was a quadratic. Applying the quadratic formula I got my two angles mentioned previously.

You mean that you put $t=\tan\theta$ to save typing or something? You needed to say when you did it. You must have divided through by cosine someplace then.

Fair enough though.

 

[Kot]

Thank you for all of the advice. I understand why my equations were unclear (still a beginner at using LaTeX). I will be sure to properly format equations in the future so they are easier to read and understand. As for the problem, I managed to finish it earlier today.

 

[Simon Bridge]

Yeah, well done :D