Question

question 28
1 pts
a ball is dropped from rest from an initial height of 9.4 meters from the ground. determine the total time for the ball to fall to the ground.
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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 ball is dropped from rest, \(v_0 = 0\). So the equation simplifies to \(h=\frac{1}{2}gt^{2}\), where \(h = 9.4\space m\) and \(g = 9.8\space m/s^{2}\).

Step2: Solve for \(t\)

From \(h=\frac{1}{2}gt^{2}\), we can solve for \(t\). First, rewrite the equation as \(t^{2}=\frac{2h}{g}\). Substitute \(h = 9.4\space m\) and \(g = 9.8\space m/s^{2}\): \(t^{2}=\frac{2\times9.4}{9.8}=\frac{18.8}{9.8}\approx1.918\). Then take the square - root of both sides: \(t=\sqrt{1.918}\approx1.39\space s\).

Answer:

\(1.39\space s\)