Question

a ball takes 0.5 s to fall to the floor from a horizontal launch. what is the height of the launch point? use g = 10 m/s².

Explanation:

Step1: Identify the formula

We use the free - fall formula $h = v_0t+\frac{1}{2}gt^{2}$. Since the ball is launched horizontally, the initial vertical velocity $v_0 = 0$. So the formula simplifies to $h=\frac{1}{2}gt^{2}$.

Step2: Substitute the values

Given $g = 10\ m/s^{2}$ and $t = 0.5\ s$. Substitute these values into the formula: $h=\frac{1}{2}\times10\times(0.5)^{2}$.

Step3: Calculate the result

First, calculate $(0.5)^{2}=0.25$. Then, $\frac{1}{2}\times10\times0.25 = 1.25\ m$.

Answer:

1.25 m