Given is the acceleration due to gravity = -32 feet/sec².
Final velocity of stone when hitting the ground = 120 feet/sec.
Part A: We know acceleration = derivative of velocity.
a = v' = -32 feet/sec²
∫ a = ∫ v' = ∫ -32
∫ dv = ∫ -32 dt
v = -32t + u.
where u is the initial velocity.
v is the velocity at time t, so v(t) = -32t + u.
The initial velocity is zero when the stone was dropped.
Part B: We know velocity = derivative of distance.
v = h' = -32t + u
∫ v = ∫ h' = ∫ (-32t + u)
∫ dh = ∫ (-32t + u) dt
h = -16t² + ut + C.
where C is a constant of integration and initial height when stone is dropped.
h is the height of the stone at time t, so h(t) = -16t² + ut + C.
Part C: Given final velocity v = 120 feet/sec, and we have v = -32t + u.
initial velocity, u = 0.
So it means -120 = -32t + 0.
t = -120/-32 = 15/4 seconds.
When the stone hit the ground, height h = 0 and time t = 15/4 seconds.
So it means 0 = -16·(15/4)² + 0·(15/4) + C
0 = -16·(225/16) + C
0 = -225 + C
C = 225 feet.
Hence, the height of cliff is 225 feet.