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

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

explanation:

wheres the pic