On another planet, a marble is released from rest at the top of a high cliff. It falls 4.00 m in the first 1 s of its motion. Through what additional distance does it fall in the next 1 s

Distance (h1) = 12 m

Explanation:

Given:

Height of cliff (h) = 4m

Time (t) = 1s

Intial velocity (u) = 0m/s

Computation:

Distance (h) = 1/2 (g) (t) ²

4 = 1/2 (g)

g = 8m/s ²

Distance (h1) = 1/2 (g) [ (t1) ² - (t) ²]

Distance (h1) = 1/2 (8) (3)

Distance (h1) = 12 m