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

#### Chemist116

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
the sketch doesn't match the problem statement ...

#### skeeter

Math Team
$\phi = \arctan\left(\dfrac{1}{\sqrt{3}}\right) = 30^\circ$

#### skeeter

Math Team
confirmation of $\phi = 30^\circ$

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

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

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

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

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

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

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