Question

Please help 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

Answer 1

C. 125

Step-by-step 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