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Free Fall

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Question 1-4 0:12
  • Question 5 2:36
  • Question 6 3:11
  • Question 7 4:44
  • Question 8 6:16

Transcription: Free Fall

Hi everyone and welcome back to

We are going to continue our work through these practice problems from the APlusPhysics site.0003

You will find the link down below. We are going to do page 1 of the free-fall worksheet. Let us take a look.0007

The first four problems are all related to this graph problem.0013

We have a 1 kg mass dropped from rest from a height of 25 m above Earth's surface.0018

The speed of the mass was determined at 5 m intervals and recorded in the data table.0023

We have our data table over here and our graph over here.0028

The first thing we are asked to do is to mark an appropriate scale on the axis labeled height above Earth's surface.0031

What I am going to do here, since we are dropping it from a height and it is going down toward the Earth, I am going to start with my height over here at 25 m -- 20, 15, 10, 5, 0.0038

And note, you could have gone from 0 to 25, but I think it is a little more intuitive as we analyze the graph as we are going to have an increase in velocity.0054

Number 2 -- Plot the data points for speed versus height above Earth's surface.0063

To plot our data points at 25 m, our speed is 0.0068

At 20 m, it is about 9.9; at 15 m, we are at 14; at 10 m, we at about 17.1; at 5 m, 19.8; and at 0 m, we are at 22.1.0073

Our graph looks something like that as far as plotting the points go.0096

Number 3 says draw the line or curve of best fit.0100

As I look at this, this does not really look like a line to me. It looks like we have a curve.0103

So I would draw a curve that looks something like that.0108

Number 4 then says, using your graph, determine the speed of the mass after it has fallen a vertical distance of 12.5 m.0115

Well, if we start at 25 m and it has fallen 12.5 m, that means its height above Earth's surface must be 25 - 12.5 or 12.5 m right there.0124

I go up to that point on the graph. That is what we are looking for and that looks to me like we are right at about 16 m, so the speed after is has fallen that far is going to be about 16 m/s.0136

Pretty straight forward for a graphing problem.0152

Looking at Number 5 -- A rock is dropped from a bridge. What happens to the magnitude of the acceleration and the speed of the rock as it falls?0156

While it is in free-fall, we are neglecting air resistance, the acceleration remains constant.0165

The acceleration is going to be 9. 8 m/s2 down, but of course we all know that the longer the rocks are in the air, the faster it is going to be going.0170

We are looking for acceleration remaining constant and speed increasing and that will be Answer 4, acceleration remains the same and speed increases.0180

Number 6 -- A rock falls from rest -- V-initial = 0 -- a vertical distance of 0.72 m -- δy = 0.72 m and we are going to call down the positive y-direction to the surface of a planet in a time of 0.63 s.0192

Find the magnitude of the acceleration due to gravity on the planet.0211

We are looking for (A), so we need the kinematic equation that has these -- V-initial, our δy, a, and t.0215

So I would probably use δy = V-initial(t) + 1/2ayt2, where this V-initial = 0, so that term is going to go away and δy = 1/2at2.0224

Let us solve for the acceleration.0243

Acceleration is going to be 2δy/t2, so I can substitute in my values, 2 × δy (0.72 m)/0.63 s2 (time)...0245

...and I come up with an acceleration when I do this of 2 × 0.72/0.632 of about 3.6 m/s2 -- Answer Number 3.0262

Number 7 -- A ball is thrown straight downward with a speed of 0.5 m/s from a height of 4 m.0283

What is the speed of the ball 0.7 s after it is released?0291

Well, if it is going downward, we will call that the positive y-direction.0295

Its initial velocity is 0.5 m/s and it is thrown from a height of 4 m, but it does not really tell us how far it has gone though, so I am not going to write any givens down for that yet.0299

What is the speed of the ball?0315

We are looking for V-final 0.7 s after it is released and of course our acceleration must be 9.8 m/s2, positive because we called down positive.0316

If I want my final velocity, V-final is just V-initial plus our acceleration times our time.0331

That 4 m does not have anything to do with the problem it is asking us.0338

This implies then that our final velocity will be (V-initial) 0.5 m/s + 9.8 m/s2 (acceleration) × 0.7 s (time)...0342 I get a velocity of 0.5 + 9.8 × 0.7 or 7.36 m/s or our best answer is 7.4 m/s.0355

Last one here -- An astronaut drops a hammer from 2 m above the surface of the moon, so initial velocity, if it is being dropped is 0.0377

If the acceleration due to gravity on the moon is 1.62 m/s2, calling down the positive y-direction again, how long will it take for the hammer to fall to the moon's surface?0386

It is going to travel a distance of 2 m, so we know initial velocity, we know acceleration, we know time, and we know the displacement.0398

I would probably use δy = V-initial × time + 1/2at2 and again V-initial is 0, so that term becomes 0 and δy is just 1/2at2...0410

...let us solve for (t) and that means t2 will be 2δy/a, therefore (t) will be the square root of 2δy/a.0427

Now let us substitute in our values -- that is 2 × 2 m (δy)/1.62 m/s2 (acceleration) and we want the square root of all of that...0440

...which implies then that our time is going to be about 1.57 s or our best answer over here is Number 3, 1.6 s.0453

That finishes up page 1 of this worksheet. We will be back soon everyone. Thanks and make it a great day!0470