two players are playing table tennis. player a hits the ball at a height of 1.24 m above the edge of the…

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.

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.

Answer

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).