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

Jun 2017
399
6
Lima, Peru
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?.
 

skeeter

Math Team
Jul 2011
3,365
1,856
Texas
the sketch doesn't match the problem statement ...
 

skeeter

Math Team
Jul 2011
3,365
1,856
Texas
$\phi = \arctan\left(\dfrac{1}{\sqrt{3}}\right) = 30^\circ$
 

skeeter

Math Team
Jul 2011
3,365
1,856
Texas
confirmation of $\phi = 30^\circ$

at $h_{max} = 2.5$, $v_y = 0$ ...

$0^2 = v_0^2 \sin^2{\phi} - 2g \Delta y \implies v_0^2 \sin^2{\phi} = 50$

$\Delta x = v_0 \cos{\phi} \cdot t \implies t = \dfrac{5\sqrt{3}}{v_0 \cos{\phi}}$

$\Delta y = v_0 \sin{\phi} \cdot t - \dfrac{1}{2}gt^2$

$2.5 = \dfrac{v_0 \sin{\phi}}{v_0 \cos{\phi}} \cdot 5\sqrt{3} - 5\left(\dfrac{5\sqrt{3}}{v_0 \cos{\phi}}\right)^2 = 5\sqrt{3} \cdot \tan{\phi} - \dfrac{15}{2} \cdot \dfrac{50}{v_0^2 \cos^2{\phi}}$

$0 = -2.5 + 5\sqrt{3} \cdot \tan{\phi} - \dfrac{15}{2} \cdot \tan^2{\phi}$

$\tan{\phi} = \dfrac{-5\sqrt{3}}{-15} = \dfrac{\sqrt{3}}{3} \implies \phi = 30^\circ$