A football kicker can give the ball an initial speed of 25 m/s. What are the (a) least and (b) greatest elevation angles at which he can kick the ball to score a field goal from a point 50 min front of goalposts whose horizontal bar is 3.44 mabove the ground?

It is given that,

The initial speed of ball is$v_0=25\mathrm{m}/\mathrm{s}$

height range of ball is, 50 m

the height of goal post is 3.44m.

As ball is hit by kicker with initial velocity and goal post has certain height. Solve this problem with quadratic equation and kinematic equation and find time by considering horizontal distance. Using that time, vertical height can be found.

Required formula is as follow:

$s=v_{0}t+\frac{1}{2}a{t}^2$ (i)

Resolving the initial velocity into two components, vertical and horizontal components will beandrespectively.

The horizontal distance travelled by a ball is,

$$\begin{gathered} s=v_{0}t+\frac{1}{2}a{t}^2 \\ x=\left(v_{0}cos\theta \right)t \\ t=\frac{x}{v_0\cos \theta }\left(2\right) \end{gathered}$$

The vertical distance travelled by ball is,

$$\begin{gathered} s=v_{0}t+\frac{1}{2}a{t}^2 \\ y=\left(v_0\sin \theta \right)t=\frac{1}{2}g{t}^2\left(3\right) \end{gathered}$$

Substituting equation (ii) in equation (iii),

$$\begin{gathered} y=\left(v_0\sin \theta \right)\left(\frac{x}{v_0\cos \theta }\right)-\frac{1}{2}g{\left(\frac{x}{v_0\cos \theta }\right)}^2 \\ y=\left(\tan \theta x\right)-\left(\frac{g{x}^2}{2v_0^2}\right)\left(\frac{1}{\cos {\theta }^2}\right) \\ y=\left(\tan \theta x\right)-\left(\frac{g{x}^2}{2v_0^2}\right)(1+\tan {\theta }^2) \\ y=\left(\tan \theta x\right)-\frac{1}{2}\left(\frac{g{x}^2}{v_0^2}\right)+\frac{1}{2}\left(\frac{g{x}^2}{v_0^2}\right)\tan {\theta }^2 \\ 0=-\left(\tan \theta x\right)+\frac{1}{2}\left(\frac{g{x}^2}{v_0^2}\right)+\frac{1}{2}\left(\frac{g{x}^2}{v_0^2}\right)\tan {\theta }^2-y \\ 0=-\left(\tan \theta x\right)+\frac{1}{2}\left(\frac{g{x}^2}{v_0^2}\right)+\frac{1}{2}\left(\frac{g{x}^2}{v_0^2}\right)\tan {\theta }^2+y \\ 0=\frac{1}{2}\left(\frac{g{x}^2}{v_0^2}\right)\tan {\theta }^2-\left(\tan \theta x\right)+y+\frac{1}{2}\left(\frac{g{x}^2}{v_0^2}\right) \end{gathered}$$

Suppose$c=\frac{1}{2}\left(\frac{g{x}^2}{v_0^2}\right)$

c = 19.6

And height of the goal post is also given as y = 3.44 m

So, above equation becomes,

$$\begin{gathered} 0=(19.6)\tan {\theta }^2-\left(\tan \theta x\right)+3.44+19.6 \\ 0=(19.6)\tan {\theta }^2-(50\tan \theta )+23.3 \end{gathered}$$

This is quadratic in tan, so, after solving, two real solutions are obtained, that is,

$$\begin{gathered} \tan \theta =1.95\mathrm{and}0.605 \\ \mathrm{Thus},\theta =63°\mathrm{and}\theta =31° \end{gathered}$$

Thus,using the given conditions, initial velocity can be resolved along two axes to find its time. By using the vertical distance formula, it can be settled into a quadratic equation. And that quadratic equation can be solved by using given parameters in problem. So that, two solution of that equation can be found and that is our answer for least and greatest angle.