Greg dropped a coin from the h meter tall building. if it takes t seconds to reach the ground, determine the position of the coin at the t/2 seconds

consider the motion of the coin in vertical direction :

Assuming down direction as positive and top of building as origin

Y₀ = initial position of displacement of the coin at the top of building = 0 m

Y = final position of displacement of the coin as it hits the ground = h

t = time taken to hit the ground

a = acceleration = acceleration due to gravity = g

v₀ = initial velocity at the top of the building = 0 m/s

using the equation

Y = Y₀ +  v₀ t + (0.5) a t²

h = 0 + 0 t + (0.5) g t²

h = (0.5) g t²                                                 eq-1

consider the motion of the coin for time "t/2"

Y₀ = initial position of displacement of the coin at the top of building = 0 m

Y' = final position of the coin after time "t/2"

t' = time of travel = t/2

a = acceleration = acceleration due to gravity = g

v₀ = initial velocity at the top of the building = 0 m/s

using the equation

Y' = Y₀ +  v₀ t' + (0.5) a t'²

Y' = 0 + 0 t + (0.5) g (t/2)²

Y' = (0.5) gt²/4

using eq-1

Y' = h/4                           since h = (0.5) g t²

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

attach the statements so i can answer : )