Question

Jason hits volleyball so that it moves with an initial velocity of 6.0 m/s straight upward. If the volleyball starts 2.0 m above the floor, how long will it be in the air before it strikes the floor? Assume that Jason is the last player to touch the ball before it hits the floor.

Answer 1

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:

$y=y_{o}+V_{oy} t +\frac{gt^{2}}{2}$   (1)

Where:

$y_{o}=2 m$  is the initial height of the volleyball

$y=0$  is the final height of the volleyball (when it finally strikes the floor)

$V_{oy}=6 m/s$ is the volleyball's initial velocity

$t$ is the time the volleyball is in the air

$g=-9.8m/s^{2}$  is the acceleration due gravity (always directed downwards)

Rewritting (1) with the given conditions:

$\frac{gt^{2}}{2} + V_{oy} t + y_{o}=0$   (2)

$-\frac{9.8 m/s^{2}}{2}t^{2} + 6 m/s t + 2 m=0$  

$-4.9 m/s^{2}t^{2} + 6 m/s t + 2 m=0$   (3)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$, which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (4)

Where:

$a=-4.9 m/s^{2}$

$b=6 m/s$

$c=2 m$

Substituting the known values:

$t=\frac{-6 \pm \sqrt{6^{2}-4((-4.9)(2)}}{2(-4.9)}$ (5)

Solving (5) we find the positive result is:

$t=1.497 s$