Question

a volleyball is jump - served at 3.5 m above the ground horizontally at a speed of 18.5 m/s. the volleyball hits the floor on the other court. what is the magnitude of the final velocity of the volleyball when it hits the ground?

Explanation:

Step1: Analyze vertical motion

The vertical motion is a free - fall motion. The initial vertical velocity \(v_{0y}=0\ m/s\), the acceleration \(a = g=9.8\ m/s^{2}\), and the vertical displacement \(y - y_{0}=- 3.5\ m\) (taking downwards as negative). Use the kinematic equation \(v_{y}^{2}=v_{0y}^{2}+2a(y - y_{0})\).
Substitute the values: \(v_{y}^{2}=0 + 2\times(- 9.8)\times(-3.5)\)
\(v_{y}^{2}=68.6\)
\(v_{y}=\sqrt{68.6}\approx8.28\ m/s\)

Step2: Analyze horizontal motion

In horizontal motion (no - acceleration, \(a_x = 0\)), the horizontal velocity remains constant. So \(v_{x}=v_{0x}=18.5\ m/s\)

Step3: Calculate the magnitude of the final velocity

Use the Pythagorean theorem \(v=\sqrt{v_{x}^{2}+v_{y}^{2}}\)
Substitute \(v_{x}=18.5\ m/s\) and \(v_{y}\approx8.28\ m/s\)
\(v=\sqrt{(18.5)^{2}+(8.28)^{2}}=\sqrt{342.25 + 68.5584}=\sqrt{410.8084}\approx20.3\ m/s\)

Answer:

\(20.3\ m/s\)