Question

two players are playing table tennis. player a hits the ball at a height of 1.24 m above the edge of the table, measured from the top of the table to the bottom of the ball. the initial speed of the ball is 12.0 m s⁻¹ horizontally. assume that air resistance is negligible. the ball bounces and then reaches a peak height of 1.18 m above the table with a horizontal speed of 10.5 m s⁻¹. the mass of the ball is 2.7 g. (a) show that the time taken for the ball to reach the surface of the table is about 0.50 s.

Explanation:

Step1: Identify vertical - motion equation

The ball's vertical - motion is a free - fall motion. The initial vertical velocity \(u_y = 0\ m/s\), the acceleration \(a = g=9.8\ m/s^{2}\), and the vertical displacement \(y = 1.24\ m\). We use the equation \(y=u_y t+\frac{1}{2}at^{2}\).
Since \(u_y = 0\ m/s\), the equation simplifies to \(y=\frac{1}{2}gt^{2}\).

Step2: Solve for time \(t\)

We can re - arrange the equation \(y = \frac{1}{2}gt^{2}\) to solve for \(t\). First, multiply both sides by 2 to get \(2y=gt^{2}\). Then, \(t=\sqrt{\frac{2y}{g}}\).
Substitute \(y = 1.24\ m\) and \(g = 9.8\ m/s^{2}\) into the formula: \(t=\sqrt{\frac{2\times1.24}{9.8}}=\sqrt{\frac{2.48}{9.8}}\approx\sqrt{0.253}\approx0.50\ s\).

Answer:

The time taken for the ball to reach the surface of the table is approximately \(0.50\ s\).