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

Question

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?.

Answer 1

$tan^{-1}(\frac{1}{\sqrt{3}})$ = 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 .

Answer 2

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 $\tan^{-1} \frac{\sqrt3}{3} = 30º$.

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(*): 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.

Answer 3

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.

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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 $$ \tan\phi = 2\frac hk = 2\frac{2.5}{5\sqrt3} = \frac{1}{\sqrt3}.$$

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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 $0 = ak^2 + h,$