Question

supposed you gently push a ball off of a table top, 0.9 m above the floor. how long (in seconds) does it take the ball to reach the floor?
a 0.183 s
b 0.428 s
c 0.0918 s
d 8.83 s

Explanation:

Step1: Identify the motion type

The ball is in free - fall motion in the vertical direction (since it's pushed gently, the initial vertical velocity \(v_{0y} = 0\ m/s\)). The vertical displacement \(y=- 0.9\ m\) (taking downwards as negative or we can take upwards as positive and then \(y = 0.9\ m\) and acceleration \(a = g=9.8\ m/s^{2}\)), and the acceleration \(a = g = 9.8\ m/s^{2}\) (acceleration due to gravity). The kinematic equation for vertical displacement in free - fall is \(y=v_{0y}t+\frac{1}{2}at^{2}\). Since \(v_{0y} = 0\ m/s\), the equation simplifies to \(y=\frac{1}{2}gt^{2}\) (if we take downwards as positive, \(y = 0.9\ m\) and \(a = g\); if we take upwards as positive, \(y=- 0.9\ m\) and \(a=-g\), but the result for \(t\) will be the same).

Step2: Solve for time \(t\)

From \(y=\frac{1}{2}gt^{2}\), we can re - arrange the formula to solve for \(t\). We get \(t=\sqrt{\frac{2y}{g}}\). Substitute \(y = 0.9\ m\) and \(g = 9.8\ m/s^{2}\) into the formula:
\(t=\sqrt{\frac{2\times0.9}{9.8}}=\sqrt{\frac{1.8}{9.8}}=\sqrt{0.18367}\approx0.428\ s\)

Answer:

B. 0.428 s