Answer:
C. Increasing the vertical height from which the ball is thrown.
Explanation:
In this case, we understand that ball experiments a parabolical motion, which is the combination of horizontal uniform motion and vertical uniform accelerated motion, due to gravity. Equations of motion for the ball are described below:
[tex]x = x_{o}+v_{o,x}\cdot t[/tex] (Eq. 1)
[tex]y = y_{o} + v_{o,y}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}[/tex] (Eq. 2)
Where:
[tex]x_{o}[/tex], [tex]x[/tex] - Initial and final horizontal positions, measured in meters.
[tex]y_{o}[/tex], [tex]y[/tex] - Initial and final vertical positions, measured in meters.
[tex]v_{o,x}[/tex], [tex]v_{o,y}[/tex] - Initial horizontal and vertical velocities, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
Now we get (Eq. 2) and solve the expression for the time:
[tex]\frac{1}{2}\cdot g\cdot t^{2}+v_{o,y}\cdot t + (y_{o}-y) = 0[/tex]
[tex]t = \frac{-v_{o,y}\pm\sqrt{v_{o,y}^{2}-2\cdot g \cdot (y_{o}-y)}}{g}[/tex]
If [tex]v_{o,y} > 0[/tex], [tex]g < 0[/tex] and [tex]y_{o} -y < 0[/tex], then maximum time occurs when:
[tex]t = \frac{-v_{o,y}+ \sqrt{v_{o,y}^{2}-2\cdot g \cdot (y_{o}-y)}}{g}[/tex]
And,
[tex]v_{o,y}^{2}-2\cdot g\cdot (y_{o}-y)\geq 0[/tex].
If [tex]y_{o}[/tex] is increased, then [tex]y_{o} - y[/tex] goes closer to zero and [tex]v_{o,y}^{2}-2\cdot g\cdot (y_{o}-y)[/tex] becomes greater and time increased. Hence, we conclude that correct answer is C.
