Question

a flowerpot falls 25.0m towards a pedestrian. how long does the pedestrian have to move out of the way? (write your answer in 3 significant figures)

Explanation:

Step1: Identify the kinematic equation

The kinematic equation for vertical motion under gravity is \(h = v_0t+\frac{1}{2}gt^2\). Since the flowerpot is dropped (\(v_0 = 0\)), the equation simplifies to \(h=\frac{1}{2}gt^2\).

Step2: Solve for \(t\)

We can rearrange \(h=\frac{1}{2}gt^2\) to find \(t\). First, multiply both sides by \(2\) to get \(2h = gt^2\). Then, divide both sides by \(g\) (\(g = 9.8\ m/s^2\)): \(t^2=\frac{2h}{g}\). Taking the square - root of both sides gives \(t=\sqrt{\frac{2h}{g}}\).
Substitute \(h = 25.0\ m\) and \(g = 9.8\ m/s^2\) into the formula: \(t=\sqrt{\frac{2\times25.0}{9.8}}\).
Calculate \(\frac{2\times25.0}{9.8}=\frac{50}{9.8}\approx5.102\). Then \(t=\sqrt{5.102}\approx2.26\ s\).

Answer:

\(2.26\ s\)