Question

1. an object is released from rest on a newly discovered planet that has no atmosphere. the object falls freely for 3.0 meters in the first second. what is the magnitude of the acceleration due to gravity on this planet? (a) 1.5 m s² (b) 3.0 m/s² (c) 6.0 m/s² (d) 12.0 m/s²

Explanation:

Step1: Write the kinematic equation

The kinematic equation for displacement \(s = v_0t+\frac{1}{2}at^{2}\). Since the object is released from rest, \(v_0 = 0\). So the equation simplifies to \(s=\frac{1}{2}at^{2}\).

Step2: Rearrange the equation to solve for \(a\)

We can rewrite the equation \(s=\frac{1}{2}at^{2}\) for \(a\) as \(a=\frac{2s}{t^{2}}\).

Step3: Substitute the given values

We are given that \(s = 3.0\space m\) and \(t = 1\space s\). Substituting these values into the formula \(a=\frac{2\times3.0}{1^{2}}\).

$$a=\frac{6}{1}=6.0\space m/s^{2}$$

Answer:

C. \(6.0\space m/s^{2}\)