Answer: 1.497 s
Explanation:
This situation is related to projectile motion or parabolic motion, in which the initial velocity of the volleyball has only y-component, since it was hit straight upward.
Being the main equation as follows:
[tex]y=y_{o}+V_{oy} t +\frac{gt^{2}}{2}[/tex] (1)
Where:
[tex]y_{o}=2 m[/tex] is the initial height of the volleyball
[tex]y=0[/tex] is the final height of the volleyball (when it finally strikes the floor)
[tex]V_{oy}=6 m/s[/tex] is the volleyball's initial velocity
[tex]t[/tex] is the time the volleyball is in the air
[tex]g=-9.8m/s^{2}[/tex] is the acceleration due gravity (always directed downwards)
Rewritting (1) with the given conditions:
[tex]\frac{gt^{2}}{2} + V_{oy} t + y_{o}=0[/tex] (2)
[tex]-\frac{9.8 m/s^{2}}{2}t^{2} + 6 m/s t + 2 m=0[/tex]
[tex]-4.9 m/s^{2}t^{2} + 6 m/s t + 2 m=0[/tex] (3)
This is a quadratic equation (also called equation of the second degree) of the form [tex]at^{2}+bt+c=0[/tex], which can be solved with the following formula:
[tex]t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/tex] (4)
Where:
[tex]a=-4.9 m/s^{2}[/tex]
[tex]b=6 m/s[/tex]
[tex]c=2 m[/tex]
Substituting the known values:
[tex]t=\frac{-6 \pm \sqrt{6^{2}-4((-4.9)(2)}}{2(-4.9)}[/tex] (5)
Solving (5) we find the positive result is:
[tex]t=1.497 s[/tex]
