Me with this question
a body is dropped from a height h from the state of rest. it covers a distance of 9h/25 in the last second. what is the height from which the body falls? (in meters)
a. 12.5
b. 1.25
c. 125
d. zero

C. 125

Explanation:

Given:

y₀ = h m

v₀ = 0 m/s

a = -10 m/s²

@ t − 1, y = 9/25 h m

@ t, y = 0 m

Find: h

Use constant acceleration equation:

y = y₀ + v₀ t + ½ at²

@ t:

0 = h + (0) (t) + ½ (-10) t²

0 = h − 5 t²

h = 5 t²

@ t − 1:

9/25 h = h + (0) (t − 1) + ½ (-10) (t − 1)²

9/25 h = h − 5 (t − 1)²

9/25 h = h − 5 (t² − 2t + 1)

9/25 h = h − 5 t² + 10 t − 5

0 = 16/25 h − 5 t² + 10 t − 5

Substitute:

0 = 16/25 (5 t²) − 5 t² + 10 t − 5

0 = 16/25 t² − t² + 2 t − 1

0 = -9/25 t² + 2 t − 1

0 = 9 t² − 50 t + 25

0 = (t − 5) (9t − 5)

t = 5/9 or 5

Since t > 1, t = 5.

Solving for h:

h = 5 t²

h = 125

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!