Question

On another planet, a marble is released from rest at the top of a high cliff. It falls4.00m in the first 1 s of its motion. Through what additional distance does it fall in the next 1s ? (a) 4.00m (b)8.00 m (c) 12.0 m (d) 16.0m (e) 20.0m

Short answer

So, the correct answer is option (c) 12.0 m.

Step-by-step solution

Equation of freely falling body

Vertical displacement of a freely falling body is given by the relation,$\mathit{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{a}}_{\mathbf{g}}{\mathit{t}}^{\mathbf{2}}$ , where Planet $a_g=$acceleration.

Explanation for correct answer

In first case, (First 1s)

Travels 4.00 m in the first 1s.

$$\begin{gathered} 4.00\text{m}=\frac{1}{2}{a}_g(1\text{s}{)}^2 \\ {a}_g=8.00m/s^2 \end{gathered}$$

In second case,(after 2s )

$$\begin{gathered} x_f=\frac{1}{2}(8.00m/s^2)(2\text{s}{)}^2 \\ =16\text{m} \end{gathered}$$

Hence, after 2seconds of travel from rest, the change in position is 16.00 m.

So, the change in position between 1s and 2s is

$16.00 m-4.00\text{m}=12 m$.

Hence, the answer is 12 m.