Question

Jason hits a volleyball so that it moves with an initial velocity of 6.0 m/s

straight upward. If the volleyball starts from 2.0 m above the floor,
how long will it be in the air before it strikes the floor?

Answer 1

Approximately
`1.5\; {\rm s}`.

(Assumptions:
`g = 9.81\; {\rm m\cdot s^(-2)}`; air resistance on the volleyball is negligible.)

Step-by-step explanation:

Under the assumptions, acceleration of the volleyball would be
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}` during the entire flight. (This value is negative since the ball is accelerating downwards- toward the ground.)

By the time the volleyball hits the ground, the volleyball would be at a position
`2.0\; {\rm m}` below where it was launched. In other words, the (vertical) displacement of the volleyball during the entire flight would be
`x = (-2.0)\; {\rm m}`. (Negative since the ball is below where it was launched.)

Apply the SUVAT equation
`(v^(2) - u^(2)) = 2\, a\, x` to find the velocity of the volleyball right before hitting the ground. In this equation:

• 
  `v` is the velocity of the volleyball right before hitting the ground,
• 
  `u = 6.0\; {\rm m\cdot s^(-1)}` is the initial velocity of the volleyball,
• 
  `a = (-9.81)\; {\rm m\cdot s^(-2)}` is the acceleration of the volleyball, and
• 
  `x = (-2.0)\; {\rm m\cdot s^(-1)}` is the displacement of the volleyball during the flight.

Rearrange this equation and solve for the velocity right before landing,
`v`. Note that because
`v\!` is raised to the power of
`2` in
`(v^(2) - u^(2)) = 2\, a\, x`, both
`v = \sqrt{u^(2) + 2\, a\, x}` and
`v = -\sqrt{u^(2) + 2\, a\, x}` could satisfy this equation. However,
`v\!\!` needs to be negative since the volleyball would be travelling downwards before reaching the ground.

Therefore, right before reaching the ground, velocity of the volleyball would be:

`\begin{aligned} v &= -\sqrt{u^(2) + 2\, a\, x \\ &= -\sqrt{(6.0)^(2) + 2\, (-9.81)\, (-2.0)} \; {\rm m\cdot s^(-1)} \\ &\approx (-8.67)\; {\rm m\cdot s^(-1)}\end{aligned}`.

In other words, velocity of this volleyball has changed from
`u = 6.0\; {\rm m\cdot s^(-1)}` (upwards) to
`v \approx (-8.67)\; {\rm m\cdot s^(-1)}` (downwards) during this flight. Divide the change in the velocity
`(v - u)` by the rate of change in velocity
`a = (-9.81)\; {\rm m \cdot s^(-2)}` to find the duration of this flight:

`\begin{aligned}t &= (v - u)/(a) \\ &\approx ((-8.67) - 6.0)/((-9.81))\; {\rm s} \\ &\approx 1.5\; {\rm s}\end{aligned}`.