Question

a coin is flicked horizontally from a table that is 1.25 m high. what is the time it takes for the coin to hit the floor? use g = 10 m/s². 1/10 1 2 3 4 2.5 s 0.25 s 0.5 s 1.0 s

Explanation:

Step1: Identify the vertical - motion equation

The vertical - displacement of the coin is given by the equation $y = v_{0y}t+\frac{1}{2}gt^{2}$, where $y$ is the vertical displacement, $v_{0y}$ is the initial vertical velocity, $t$ is the time, and $g$ is the acceleration due to gravity. Since the coin is flicked horizontally, $v_{0y}=0$. The equation simplifies to $y=\frac{1}{2}gt^{2}$.

Step2: Rearrange the equation for time

We know that $y = 1.25$ m and $g = 10$ m/s². Rearranging $y=\frac{1}{2}gt^{2}$ for $t$, we get $t=\sqrt{\frac{2y}{g}}$.

Step3: Substitute the values

Substitute $y = 1.25$ m and $g = 10$ m/s² into the formula $t=\sqrt{\frac{2y}{g}}$. So, $t=\sqrt{\frac{2\times1.25}{10}}=\sqrt{\frac{2.5}{10}}=\sqrt{0.25}=0.5$ s.

Answer:

C. 0.5 s