Ted throws an object straight up in the air with initial velocity of 54 ft./s for a platform of 40 feet above the ground how all of the object for hits the ground

Part 1: find how many seconds it takes for the projectile to reach it's peak (velocity = 0):
V = V + at
0 = 54 + -32.2*t
t=1.67seconds
part 2: find how high in the air the projectile has gone (now knowing the time to peak):
H = H + V(t) + (1/2)at^2
H = 40+54(1.67)+(1/2)(-32.2)(1.67^2)
H= 85.27 feet
part 3: find out how much time it takes to fall from 85.27 feet (peak) to the ground below
H = H + V(t) + (1/2)at^2
0 = 85.27+(0)(t)+(1/2)(-32.2)(t^2)
t = 2.3 seconds
part4: combine the time to reach the peak plus the time to fall from peak to the ground:
total time = 1.67 + 2.3
total time = 4 seconds

Ithink the answer should be b.

The answer is