91Ó°ÊÓ

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

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, $\mathrm{y}-{\mathrm{y}}_0={\mathrm{V}}_{0\mathrm{y}}\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2$ …(¾±)

The velocity of a body in motion, ${\mathrm{V}}_{\mathrm{x}}=\frac{\mathrm{x}-{\mathrm{x}}_0}{\mathrm{t}}$ …(¾±¾±)

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

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

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.

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

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

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.