Question

For women’s volleyball the top of the net is 2.24above the floor and the court measures 9.0mby 9.0 m on each side of the net. Using a jump serve, a player strikes the ball at a point that is 3.0m above the floor and a horizontal distance of 8.0m from the net. If the initial velocity of the ball is horizontal, (a) what minimum magnitude must it have if the ball is to clear the net and (b) what maximum magnitude can it have if the ball is to strike the floor inside the back line on the other side of the net?

Short answer

a) The minimum initial velocity required to clear the net is$20.3\mathrm{m}/\mathrm{s}$.

b) The maximum velocity possible to strike the floor inside the back line is$21.7\mathrm{m}/\mathrm{s}$

Step-by-step solution

The given data

$$\begin{gathered} 1)\mathrm{Measurement}\mathrm{of}\mathrm{the}\mathrm{net}\mathrm{is}9.0\mathrm{m}\times 9.0\mathrm{m}. \\ 2)\mathrm{The}\mathrm{top}\mathrm{of}\mathrm{the}\mathrm{net}\mathrm{is}2.24\mathrm{above}\mathrm{the}\mathrm{floor}. \\ 3)\mathrm{Player}\mathrm{strikes}\mathrm{the}\mathrm{ball}\mathrm{at}\mathrm{height}3.0\mathrm{m} \\ 4)\mathrm{Horizontal}\mathrm{distance}\mathrm{from}\mathrm{the}\mathrm{net}8.0\mathrm{m} \end{gathered}$$

Understanding the concept of the projectile motion

Projectile motion is the motion of a particle that is launched with an initial velocity. During its flight, the particle’s horizontal acceleration is zero and its vertical acceleration is the free-fall acceleration. (Upward is taken to be a positive direction.)

We can use the concept of projectile motion for the motion of volleyball. As the motion of volleyball has only gravitational acceleration acting on it, so we can use the equations of constant acceleration.

Formulae:

The second equation of kinematic motion, $\mathrm{y}-{\mathrm{y}}_0={\mathrm{V}}_{0\mathrm{y}}\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2$ …(i)

The velocity of a body in motion, ${\mathrm{V}}_{\mathrm{x}}=\frac{\mathrm{x}-{\mathrm{x}}_0}{\mathrm{t}}$ …(ii)

(a) Calculation of the minimum initial velocity to clear the net

Considering the vertical motion, we can plug the given values in equation (i) as follows:

$$\begin{gathered} 2.24\mathrm{m}-3.0\mathrm{m}=0-\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\times {\mathrm{t}}^2 \\ \mathrm{t}=0.3938\mathrm{s} \end{gathered}$$

Now, using the horizontal motion, the velocity required is given using equation (ii) as follows:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{x}}=\frac{8.0\mathrm{m}}{0.3938\mathrm{s}} \\ =20.3\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the velocity is 20.3 m/s.

(b) Calculation of the possible maximum velocity

This part is similar to above, the difference is here y = 0, so the time is given using equation (i) as follows:

$$\begin{gathered} 0\mathrm{m}-3.0\mathrm{m}=0-\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\times {\mathrm{t}}^2 \\ \mathrm{t}=0.782s \end{gathered}$$

Again, considering the horizontal motion, we get the value of the possible maximum velocity to strike the floor using equation (ii) as follows:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{x}}=\frac{8.0\mathrm{m}+9.0\mathrm{m}}{0.782\mathrm{s}} \\ =21.7\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the velocity is 21.7 m/s.