Let t be the amunt of seconds that have passed when the height of the cannonball above the ground is the same as its horizontal fistance from the cliff.
Since the height of the cannonball above the ground is represented using the variable y and the horizontal distance from the cliff is represented using the variable x, then, the condition that the height equals the horizontal distance can be expressed as:
[tex]y=x[/tex]Replace the expressions for y and x in terms of t into the equation:
[tex]-5t^2+2t+20=2t[/tex]We obtained a quadratic equation on the variable t.
Notice that the term 2t appears in both members of the equation. Then, it can be cancelled out:
[tex]-5t^2+20=0[/tex]Solve for t²:
[tex]\begin{gathered} \Rightarrow-5t^2=-20_{} \\ \Rightarrow t^2=\frac{-20}{-5} \\ \Rightarrow t^2=4 \end{gathered}[/tex]Take the square root to solve for t:
[tex]\begin{gathered} \Rightarrow t=\pm\sqrt[]{4} \\ =\pm2 \end{gathered}[/tex]Since t must be greater or equal to 0, then the negative solution should be discarded.
Therefore, the vertical height of the cannonball equals its horizontal distance from the cliff 2 seconds after the ball is fired.
The correct choice is option B) 2
