# Homework Help: Question about finding height, given vi and time

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1. Apr 26, 2015

### mykosfemme

Hello all! I am prepping for my finals and going over some work I had problems with before. This one I am still struggling with finding for some reason. Any help would be appreciated!

So here is the question:

A stone is thrown horizontally with an initial speed of 20 m/s from the edge of a cliff. A stop watch measures the stone's trajectory time from the top of the cliff to the bottom to be 4.0 s. What is the height of the cliff?
The answer is anything between 78 - 78.5

At first I tried v=d/t: 20 = d/4.0 = 80 meters. This is close, but not correct.

Then I tried v^2 = vo^2 + 2ax: 0 = 400 + 2(9.8)x => x= -20.41 meters. This is also not correct.

What am I not getting?

2. Apr 26, 2015

### phinds

What would the answer be if the stone were simply dropped off the cliff?

3. Apr 26, 2015

### Staff: Mentor

Hi mykosfemme, Welcome to Physics Forums.

In future, please use the formatting template provided in the edit window when starting a thread here in the Homework sections of PF. This is a requirement of the forum rules.

Regarding your question, consider that a trajectory local to the Earth's surface has two components: horizontal and vertical. They are independent and have separate equations of motion (do you know why?). What component of the trajectory are you interested in for this problem?

4. Apr 27, 2015

### mykosfemme

Gneill - my apologies! I will stick to the format in the future.

phinds & gneill: I am not entirely sure.
I tried using v = sqrt 2gh => 10 = sqrt 19.6h
100 = 19.6h^2
5.12 = h^2
h= 2.26
And this is not correct.

I got closer using x = xo +vo +1/2 at^2:
0 = xo + 10 + 1/2*-9.8*5^2
0= xo + 10 -122.5
xo = 112.5

I am quite stuck on this problem for some reason. I know it is simple, but I cannot remember it.

5. Apr 27, 2015

### phinds

"remembering" it is a bad idea. "Understanding" it would be a lot better.

If you drop a rock off a tall building so that it travels straight down, what is the equation for the motion of the rock? Assume the top of the building is the origin for simplicity.

6. Apr 27, 2015

### Staff: Mentor

Where does the 10 come from? I can see that the 19.6 comes from 2g.

Anyway, the approach is not correct since it doesn't involve the only relevant piece of information that you are given: the time for the projectile to reach the ground.
That's better approach. Again, where does the 10 come from?

Have you identified the separate components of the motion? You seem to be attributing an initial speed to the vertical motion. Why is that? How is the projectile thrown?

So,
Step 1: Identify the initial conditions of the motion, horizontal and vertical.
Step 2: Identify relevant kinematic equations (SUVAT) that contain variables for which you have known values as well as the unknown that you wish to find.
Step 3: Choose one of the selected equations and solve for your unknown.

7. Apr 27, 2015

### mykosfemme

I'm sorry, by 10 I meant 20, as in 20 m/s. I got the value confused with another similar problem I was looking at.

1. Conditions: I know finding the horizontal component is Vcostheta and the vertical is Vsintheta. For this problem, the stone is thrown horizontally and the only vertical force acting is g. I know you can find V using sqrt Vx^2 and vy^2.

2. I haven't heard of SUVAT before. I google it and saw the kinematics equations. Solving for S, I got:
s = (u+v)t/2 => (20 + 0)5/2 => s=50
s = ut + 1/2at^2 => 100 +122.5 => s = 222.5 (using g as a)

I'm sorry if it seems I am not getting this.

8. Apr 27, 2015

### Staff: Mentor

You are mixing up parameters from the horizontal and vertical motions.

The given initial conditions are:

Horizontal: Initial velocity 20 m/s; Initial position assumed to be at x = 0; acceleration is zero.
Vertical: Initial velocity 0 m/s; Initial position is unknown height h; acceleration is g downwards.

Note in particular the initial velocity in the vertical direction! You should confirm that these initial conditions can be found from the given information in the problem statement.

9. Apr 28, 2015

### mykosfemme

Sorry for delayed reply! I'm still quite confused, but here's another attempt:

Ok, so.... horizontal: x=0, a = 0 m/s^2, vi = 20 m/s
x = xo + vot + 1/2 at^2 = > 0 = x0 +20(4) + (1/2)(0)(4)^2
x0 = -80

vertical vi = 0 m/s, h = ?, a = -9.8
x = xo + vot + 1/2at^2
x = 80 + 0(4) + 1/2(-9.8)(4)^2
x= 158.4

I got the answer with 1/2(9.8)(4^2) =78.4. Why is this? What happened to the other values of the equation? Do they cancel out somehow?

10. Apr 28, 2015

### phinds

You really should pay attention to post #2

11. Apr 28, 2015

### mykosfemme

I gave it a lot of consideration, but I don't know what the difference in setting up the equation would be. I'm really not trying to be difficult, I'm just very weak with this type of problem for some reason.

12. Apr 28, 2015

### Staff: Mentor

No, you're given x0 = 0. It's the starting position (in the horizontal direction) of the stone. It's where the stone is launched from. What you might be interested in is how far away (in the x-direction) the stone is at time t=4s simply because that's the time that the stone hits the ground according to the problem statement. Although, the question doesn't actually ask for that information and is not required to answer the problem.
Why have you imported a horizontal position, 80 (which occurs at time t=4s), into the vertical motion as an initial position? Horizontal and vertical motions are ENTIRELY independent. Do not move values from one to another; it's like taking values from an unrelated problem to use in some other problem. The only thing that links the separate horizontal and vertical motions is the time parameter, t.

For separate components of the motion you should choose different variable names in order to keep them distinct and clear. Most often y or z is used for motion in the vertical direction. So write the vertical equation of motion as:

y = yo + vot + (1/2)at2

You're looking to solve for yo, the initial height which we've previously designated "h".
That gives the correct result because it comes from solving the equation of motion in the vertical direction for the initial position with all the correct values inserted. Look closely at the problem statement. At time 0 the stone begins its vertical journey at some unknown height yo = h that you wish to find. You know that it falls for time 4.0s at which instant it will have reached the ground (y = 0). So you wish to solve the equation:

y = yo + vot + (1/2) a t2

where yo is the unknown height h, vo = 0, and a = -g. That yields:

0 = h - (1/2) g t2

Solve for h.

Last edited: Apr 28, 2015
13. Apr 28, 2015

### phinds

Well, you may have given it a lot of consideration but you didn't do what it asks. If you had, you would have the answer by now. Gneill and I have told you the same thing repeatedly.

14. Apr 29, 2015

### CWatters

Hint: It's almost a trick problem in that you are given more info than you need. Hope I'm not giving too much away if I say it's not really a projectile motion problem.

As other have said.. Answer post #2. Post #2 was a really strong hint that you missed.