## Fundamentals of Physics Extended (10th Edition)

$\tan \alpha _{\max }=1,28\left( 1+\sqrt {0,385-0,031\times 3,44}\right) \approx 1,96\Rightarrow \alpha _{\max }\approx 62,97^{0}$
Let's assume that initial speed of the ball is $v$ and the initial elevation angle is $α$ relative to then horizontal line. Let's write a equation that describes height of the ball as function of time: $H=v_{0}t\sin \alpha -\dfrac {gt^{2}}{2}\left( 1\right)$ And the horizontal distance ball traveled as a function of time: $x=v_{0}t\cos \alpha \left( 2\right)$ From (1) and (2) we get $H(x)=x\tan \alpha -\dfrac {gx^{2}}{2v^{2}_{0}\cos ^{2}\alpha }(3)$ $\cos ^{2}\alpha =\dfrac {1}{1+\tan ^{2}\alpha }\left( 4\right)$ So from (3) and (4) we get $\dfrac {gx^{2}}{2v^{2}_{0}}\tan ^{2}\alpha -x\tan \alpha +\left( H+\dfrac {gx^{2}}{2v^{2}_{0}}\right) =0$ İf solve this equation for $tanα$ we get $\tan \alpha =\dfrac {v_{0}^{2}}{gx}\left( 1\pm \sqrt {\left( 1-\dfrac {2gH}{v_{0}^{2}}-\dfrac {g^{2}x^{2}}{v_{0}^{4}}\right) }\right) (5)$ There are 2 solution for this equation $\tan \alpha_{max} =\dfrac {v_{0}^{2}}{gx}\left( 1+ \sqrt {\left( 1-\dfrac {2gH}{v_{0}^{2}}-\dfrac {g^{2}x^{2}}{v_{0}^{4}}\right) }\right)$ $\tan \alpha _{min}=\dfrac {v_{0}^{2}}{gx}\left( 1- \sqrt {\left( 1-\dfrac {2gH}{v_{0}^{2}}-\dfrac {g^{2}x^{2}}{v_{0}^{4}}\right) }\right)$ So if we put given numbers given $x=50m;v_{0}=25\dfrac {m}{s};g=9,8\dfrac {m}{s^{2}};H \geq 3.44m$( in order to score a goal H must be equal or greater than 3.44 m ) we get $\tan \alpha _{\max }=1,28\left( 1+\sqrt {0,385-0,031H}\right)$ in order to calculate $α_{max}$ we need to find maximum value of $tanα_{max}$ and in order $tanα_{max}$ to be maximum H should be minimum Minimal value H can be is $3.44 m$ So $\tan \alpha _{\max }=1,28\left( 1+\sqrt {0,385-0,031\times 3,44}\right) \approx 1,96\Rightarrow \alpha _{\max }\approx 62,97^{0}$