!  a stone was dropped off a cliff and hit the ground with a speed of 120 ft/s. you may assume that the acceleration due to gravity is −32 ft/s2.(a) use antiderivatives to find a formula for the velocity at time t. hint: since the stone is dropped, what is the initial velocity? b) use antiderivatives to find a formula for the height of the stone at time t. your formula may contain a constant c. c) what is the height of the cliff?

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.

idk

step-by-step explanation: i just don't