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

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s