On another planet, a morble 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?
(a) 4.00 m
(b) 8.00 m
(c) 12.0 m
(d) 16.0 m
(e) 20.0 m

option  c is correct

additional distance is 12 m

Explanation:

given data

height = 4 m

time = 1 s

to find out

additional distance in next 1 s

solution

we will apply here formula for height

height = 1/2 × g×t²       1

put here value height = 4 and t = 1 s

4 = 1/2 × g×(1)²

g = 8 m/s²

and

for time next 1 sec

from equation 1

height = 1/2 × g×t²  

put here t = 2 s so

h2 - h1 = 1/2 × g×(2²  - 1² )

h2 - h1 = 1/2 × 8×(2²  - 1² )

h2 - h1 = 4 ( 4-1)

h2 - h1 = 12 m

so option  c is correct

additional distance is 12 m