# How to find the launch angle for a ball from a fixed point where it is known the maximum height?

The problem is as follows:

From point indicated in the picture, a football player is about to kick a ball giving the ball a velocity of $$v_{o}$$. The projectile collisions with the crossbar on point $$A$$ as shown in the picture. Find the launch angle. It is known that the highest point of the trajectory is $$2.5\,m$$ and the horizontal distance from the ball to the crossbar is $$5\sqrt{3}$$.

The alternatives given are as follows:

$$\begin{array}{ll} 1.&15^{\circ}\\ 2.&30^{\circ}\\ 3.&37^{\circ}\\ 4.&45^{\circ}\\ 5.&\arctan{0.28}\\ \end{array}$$

What I've attempted here was to use the equation for the trajectory of the projectile as shown below:

$$y=x\tan\phi-\frac{1}{2}g\left(\frac{x}{v_{o}\cos\phi}\right)^2$$

Since it mentions that the highest point of this trajectory is $$2.5\,m$$ then I'll obtain from the above equation a relationship.

Using the first derivative with respect of $$x$$ I'm getting:

$$y'=\tan\phi-g\frac{x}{v_{o}^2\cos^2\phi}$$

Equating this to zero:

$$0=\tan\phi-g\frac{x}{v_{o}^2\cos^2\phi}$$

$$x=\frac{v_{o}^2\cos^2\phi\tan\phi}{g}=\frac{v_o^2\sin\phi\cos\phi}{g}$$

Then:

Inserting this in the equation for the trajectory I'm getting:

$$y=\left(\frac{v_o^2\sin\phi\cos\phi}{g}\right)\tan\phi-\frac{1}{2}g\left(\frac{\frac{v_o^2\sin\phi\cos\phi}{g}}{v_{o}\cos\phi}\right)^2$$

$$y=\left(\frac{v_o^2\sin\phi\cos\phi}{g}\right)\frac{\sin\phi}{\cos\phi}-\frac{1}{2}g\left(\frac{\frac{v_o^2\sin\phi\cos\phi}{g}}{v_{o}\cos\phi}\right)^2$$

$$y=\left(\frac{v_o^2\sin\phi\cos\phi}{g}\right)\frac{\sin\phi}{\cos\phi}-\frac{1}{2}g\left(\frac{v_o\sin\phi}{g}\right)^2$$

$$y=\frac{v_{o}^2\sin^2\phi}{g}-\frac{v_o^2\sin^2\phi}{2g}=\frac{v_o^2\sin^2\phi}{2g}$$

Then:

$$x=\frac{v_o^2\sin\phi\cos\phi}{g}$$

Since what it is given is the coordinates for each then this is reduced to:

$$5\sqrt{3}=\frac{v_o^2\sin\phi\cos\phi}{g}$$

$$2.5=\frac{v_o^2\sin^2\phi}{2g}$$

$$v_{o}^2\sin\phi=\frac{5g}{\sin\phi}$$

This is inserted in the above equation and becomes into:

$$5\sqrt{3}=\frac{v_o^2\sin\phi\cos\phi}{g}$$

$$5\sqrt{3}=\frac{\frac{5g}{\sin\phi}\cos\phi}{g}$$

$$5\sqrt{3}=\frac{\frac{5g}{\sin\phi}\cos\phi}{g}$$

$$5\sqrt{3}=\frac{5\cos\phi}{\sin\phi}$$

$$\tan\phi=\frac{1}{\sqrt{3}}$$

Therefore this becomes into:

$$\phi=\tan^{-1}\left(0.5777\right)$$

But this does not appear in any of the alternatives. Did I made a mistake or anything?. Can someone help me with this?.

$$tan^{-1}(\frac{1}{\sqrt{3}})$$ = 30°, so that pretty much match, to prove it make a equilateral triangle then draw a angle bisector from any vertex. Then find tan(30°).But i think you should know that .

• I don't know if my method does deserve an upvote as it seems that I arrived to the answer more or less but from looking at your answer I overlooked the fact that the right triangle $30-60-90$ does produce a tangent of $\frac{1}{\sqrt{3}}$. The part where you mention to prove it. Yes I can see it from the equilateral triangle produces two $30-60-90$ right triangles. But can you add a drawing of this as part of your answer?. It would help me to better visualize the answer and it could help others to see it as well. Jan 27, 2020 at 12:46
• You see that that the angle bisector length is $\frac{\sqrt{3}a}{2}$ and the lenth is halved and becomes $\frac{a}{2}$. (angle bisector is median as well as prendicular). Jan 27, 2020 at 18:38

Here is an alternative method:

The path traced by the ball is a parabola (*). If the ball's starting point is at $$(0,0)$$, the vertex form of the parabola is $$a(x-5\sqrt{3})^{2}+\frac{5}{2}$$.

Since the ball passes through $$(0,0)$$, we have $$a(-5\sqrt{3})^2 + \frac{5}{2} = 0 \Rightarrow a = -\frac{1}{30}$$.

Now differentiating, we have that $$f(x) = -\frac{1}{30} (x-5\sqrt{3})^{2}+\frac{5}{2}$$ so $$f'(x) = -\frac{1}{15}(x-5\sqrt{3})$$ and $$f'(0) = \frac{\sqrt3}{3}$$. Therefore the launch angle is $$\tan^{-1} \frac{\sqrt3}{3} = 30º$$.

(*): This is because $$g$$, the Earth's acceleration due to gravity, is measured in $$m/s^2$$, which means that the downward distance travelled by the ball increases quadratically as each second passes.

• It seems that your answer is not displaying correctly in my computer. Can you please recheck your code?. By the way, did my method validates yours?. Jan 27, 2020 at 12:48
• @ChrisSteinbeckBell Try reloading the page. That worked for me. Jan 27, 2020 at 12:50
• Do you have the latest version of your browser? If the problem persists you can post on the meta site. My method validates your method. Jan 27, 2020 at 12:51
• @TobyMak I'm using chrome and it is now displaying correctly. Reloading solved the problem. If you don't mind, can you please check this problem?. It is similar to this one but I'm still stuck. Jan 27, 2020 at 12:54

I'm going to pick a nit with the problem statement. It says the ball hits the crossbar at the highest point of the trajectory. But the free-fall trajectory that starts with the kick ends at the crossbar. So if the ball were still going up as it hit the crossbar, that would still be the highest point in its trajectory.

I think what the problem statement was intending to say was that the crossbar is at what would have been the highest point in the trajectory if the crossbar had not been there. A more concise way to say this is that the ball is traveling horizontally when it strikes the crossbar.

You interpreted the problem as I think it was intended, and did everything correctly up to the last step, which was to match the correct value of the tangent that you found with one of the angles.

Since this is multiple choice, you could just try each angle, starting with the ones whose tangents are easiest to compute. For $$30^\circ$$, you can draw this triangle:

Knowing that $$\sin(30^\circ) = \frac{BC}{AC} = \frac12,$$ you can put $$1$$ on the opposite leg and $$2$$ on the hypotenuse, then use the Pythagorean Theorem to fill in the adjacent leg. Now compute the tangent and compare to what you already have.

As a slightly different approach, you can put the equation for the parabolic trajectory into vertex form:

\begin{align} y &= x\tan\phi - \frac12 g\left(\frac{x}{v_o\cos\phi}\right)^2 \\ &= -\frac{g}{2v_o^2\cos^2\phi} \left(-\frac{2v_o^2\cos^2\phi}{g} x\tan\phi + x^2\right) \\ &= -\frac{g}{2v_o^2\cos^2\phi}\left(x^2 - \frac{2v_o^2\sin\phi\cos\phi}{g} x\right) \\ &= -\frac{g}{2v_o^2\cos^2\phi} \left(\left(x - \frac{v_o^2\sin\phi\cos\phi}{g}\right)^2 - \frac{v_o^4\sin^2\phi\cos^2\phi}{g^2}\right) \\ &= -\frac{g}{2v_o^2\cos^2\phi}\left(x - \frac{v_o^2\sin\phi\cos\phi}{g}\right)^2 + \frac{v_o^2\sin^2\phi}{2g} \\ \end{align}

That is, $$y = a(x - k)^2 + h$$ where \begin{align} a &= -\frac{g}{2v_o^2\cos^2\phi},\\ k &= \frac{v_o^2\sin\phi\cos\phi}{g},\\ h &= \frac{v_o^2\sin^2\phi}{2g}. \end{align}

Note all the common factors in $$h$$ and $$k,$$ which cancel out nicely if we take the ratio of those two parameters: $$\frac hk = \frac{\left(\frac{v_o^2\sin^2\phi}{2g}\right)} {\left(\frac{v_o^2\sin\phi\cos\phi}{g}\right)} = \frac{\tan\phi}{2}.$$

So $$\tan\phi = 2\frac hk.$$ But according to the problem statement, $$h = 2.5$$ and $$k = 5\sqrt3.$$ So $$\tan\phi = 2\frac hk = 2\frac{2.5}{5\sqrt3} = \frac{1}{\sqrt3}.$$

Note that you can also deduce the vertex form from first principles or by using other well-known properties of the trajectory.

We know the maximum height of the trajectory is $$\frac{v_o^2\sin^2\phi}{2g},$$ so that's $$h.$$ The range is $$\frac{v_o^2\sin(2\phi)}{g} = \frac{2v_o^2\sin\phi\cos\phi}{g},$$ so $$k$$ is half of that. And although you don't need $$a$$ for this problem, you can find it by setting $$x = y = 0$$ and solving for $$a$$ in $$0 = ak^2 + h,$$

• Actually this does seem to validate what I did just I overlooked the fact that the right triangle of $30-60-90$ does produce $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$. Can you please take a look into this problem I'm struggling with this. Jan 27, 2020 at 14:22
• Your calculations look fine. You asked for a diagram of the tangent so I included one. That triangle diagram is a kind of tool I still find useful. Jan 27, 2020 at 16:05