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

Answer:

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

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

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:

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