# How high is the cliff?

1. Dec 29, 2013

### Medgirl314

1. The problem statement, all variables and given/known data

An arrow is fired horizontally off a cliff. The initial speed of the arrow is 280 ft/s and the arrow lands 400 feet from the base of the cliff. How high is the cliff?

2. Relevant equations
t=x/v
y=1/2(a)(t^2)

3. The attempt at a solution

I did the horizontal component first.

xinitial=400
a=0
vinital=280 ft/s
x=vt
t=x/v
t=400/280
t=1.43 s

Vertical:
t=1.43 s
a=9.8 m/s^2
y0=0
y=?

y=1/2(a)(t^2)
y=1/2(9.8)(1.43)^2
y=9.8(2.04)^2

y=20 ft tall.

Is this correct? If so, why didn't I have to half my answer?

2. Dec 29, 2013

### SteamKing

Staff Emeritus
Your algebra is a little fuzzy when you are calculating y. Numbers seem to appear and disappear without reason.

Also, the problem distances are given in feet, and you are using g = 9.8 m/s^2. I hope you realize that feet and meters can't be used interchangeably.

3. Dec 29, 2013

### Medgirl314

I'm not sure what you mean by "numbers". Would you mind elaborating which ones? Part of it may have been done mentally.
Yes, I realize that feet and meters can't be used interchangeably, I just got used to using meters and forgot to convert. However, is converting essential for this problem? I don't see a step where the units clash, but I could just be looking over one. It looks like I have feet with feet and meters with meters throughout the problem, but again, I could be wrong. Same for the other problem. Are my answers incorrect, or were you just pointing out areas where you thought I may be unclear? Thanks again!

4. Dec 29, 2013

### SteamKing

Staff Emeritus
Specifically:

y=1/2(a)(t^2)
y=1/2(9.8)(1.43)^2 - I understand this, since you have previously solved for t

But a = 9.8 m/s^2, and if you multiply 9.8 m/s^2 by time squared, your remaining units will be meters, not feet.

y=9.8(2.04)^2 - where does 2.04 come from? What happened to the (1/2) and t = 1.43 s?

y=20 ft tall. - Again, how did you arrive at this answer from the previous step?

5. Dec 30, 2013

### Medgirl314

Ah, I see. Okay, the 2.04 came from squaring the 1.43. I just left the ^2 on, oops. I multiplied the 9.8 by the 2.04 to get 19.992, and rounded to 20. The units should have been meters. Do I half the answer? I thought so, but in an example problem my teacher did similar to this, he didn't half the answer, so I thought I may have been missing something cancel out.

Thanks again!

6. Dec 30, 2013

### PhanthomJay

You cannot just dump the 1/2 ....it is part of the equation. Also, please determine the height of the cliff in feet! You may use g = 32.2. ft/sec^2.

7. Dec 30, 2013

Medgirl314,

8. Dec 30, 2013

### Medgirl314

PhantomJay, I didn't think I could just dump the 1/2, but in a previous problem, it seemed that my teacher did, so I thought maybe it canceled out somewhere. I see it doesn't now, so I can fix the problem. Adjacent, I know you can't use feet and meters interchangeably, but I just forgot for these problems and was trying to find where the units imposed a problem. I see where now, so I'll come back to this problem after we finish the other one.

Thanks!

9. Dec 30, 2013

### Medgirl314

Is the answer 32.9 ft. ?

10. Dec 30, 2013

Right

11. Dec 30, 2013

### Medgirl314

Thanks! :D I should probably round to 33 ft for this problem. Thanks again!