10.1.2 Gravity and Vertical Motion Flashcards Preview

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Flashcards in 10.1.2 Gravity and Vertical Motion Deck (11):
1

Gravity and Vertical Motion

• The acceleration due to gravity of an object is a constant.
• Given the initial velocity and initial position of an object moving vertically, you can use the fact that the acceleration due to gravity is a constant to find the velocity and position functions.
• The velocity and position functions can be used to answer many questions about the motion of an object.

2

note

- In vertical motion, the acceleration due to gravity is a
constant 32 ft/sec 2 downward. Therefore, acceleration as a function of time can be stated as a(t) = –32.
- When studying the vertical motion of an object such as a ball, the initial vertical position is denoted y 0 . Depending on how the object is released, there are three possibilities for its initial velocity v 0 .
- Integrating the acceleration function produces the velocity function v(t) = –32t + v 0 , with v 0 replacing the constant of integration.
- Integrating the velocity function produces the position
function y(t). Notice that the constant of integration is
replaced by the initial position y 0 .
- To answer these questions you will first need to write down what you know.
- The problem statement tells you that the initial velocity was 96 ft/sec and the initial position was 0 ft.
- Use these values in the formula for vertical position.
- To find out how long the ball was in the air, you need to know when it landed, which is when its position was 0.
- There are two solutions. You can discard the solution t = 0, since that was when the ball left the machine.
- The remaining solution gives you the time the ball landed. The ball was in the air for 6 seconds.

3

note 2

- To determine the maximum height, you will need to set the derivative of the position function equal to 0.
- In other words, set the velocity function equal to 0.
- The result is the time of the maximum height. Evaluate the position function for that value of t to determine the maximum height.
- The final question asks for the velocity of the ball at the
instant it struck the ground.
- Once the ball has hit the ground it will no longer be moving. Its velocity will be 0. However, this question is asking for the velocity at the instant the ball makes contact with the ground.
- You already determined that the ball falls to ground after 6 seconds.
- Insert t = 6 into the formula for velocity. The velocity is
–96 ft/sec. Recall that the negative sign means that the ball was traveling downward. The remaining solution gives you the time the ball landed. The ball was in the air for 6 seconds.

4

A stone is dropped from a 400 foot high sea cliff at time t = 0. The acceleration due to gravity is given by a (t) = −32 ft / sec2. How many seconds after dropping the stone will it hit the water?

5 seconds

5

A ball is thrown vertically upwards from the ground with an initial velocity of 50 ft / sec. The ball is accelerated by gravity so that its acceleration is given by the function a (t) = −32 ft / sec2. What is the maximum height of the ball?

39 feet

6

A ball is thrown vertically upwards from the ground with an initial velocity of 50 ft / sec. The ball is accelerated by gravity so that its acceleration is given by the function a (t) = −32 ft / sec2 . At what velocity does the ball hit the ground?

−50 ft / sec

7

A stone is dropped from a 400 foot high sea cliff at time t = 0. The acceleration due to gravity is given by a (t) = −32. Find a formula for the velocity of the stone at any time t.

−32t

8

A bowling ball dropped out of a window hits the ground with a velocity of −128 ft / sec. How high above the ground is the window? Assume that the acceleration of the bowling ball due to gravity is −32 ft / sec2.

256 ft

9

A penny is thrown downward from a tower that is 200 feet above ground with an initial velocity of 40 ft / sec. The acceleration of gravity is −32 ft / sec2 . When does the penny hit the ground?

2.5 seconds

10

A stone is dropped from a 400 foot high sea cliff at time t = 0. The acceleration due to gravity is given by a (t) = −32 ft / sec2. What is the stone’s velocity the instant that it hits the water?

−160 feet / sec

11

A stone is dropped from a 400 foot high sea cliff at time t = 0. The acceleration due to gravity is given by a (t) = −32 ft / sec2. Find a formula for the distance the stone is above the water at any time t.

−16t ^2 + 400

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