170k views
3 votes
Help ASAP!!!!!  

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?

User Elida
by
8.6k points

1 Answer

4 votes

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.

User Riken Shah
by
8.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories