Where Does the Arrow Land? Solving for Horizontal Projectile Motion

[Medgirl314]

Homework Statement

An arrow is fired horizontally off a cliff. The initial speed of the arrow is 250 ft/s and the cliff is 22 ft high. How far from the base of the cliff does the arrow land?

Homework Equations

y=1/2a(t^2)
x=vt

The Attempt at a Solution

Vertical:

y=1/2 a(t^2)
t=(√2y)/a

t=2.12 s

Horizontal:
x=vt
x=250(2.12)
x=530 ft.

Is this answer correct? Thanks in advance!

 

[SteamKing]

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

 

[Medgirl314]

I do realize that, I just forgot to convert. I can't find anywhere where the units clash, though, so is it right anyway?

Thanks again!

 

[SteamKing]

I do realize that, I just forgot to convert. I can't find anywhere where the units clash, though, so is it right anyway?

Thanks again!

Vertical:

y=1/2 a(t^2)
t=(√2y)/a

t=2.12 s

If you have height in feet and a (or g) in m/s^2, I would say the units "clash".

 

[Medgirl314]

Ah, so would I. I didn't see that at first. But I suppose I phrased my question wrong. Did it affect the answer numerically? Is the time still right, and I just got the units wrong? Or do I need to convert for some reason?

Thanks again!

 

[adjacent]

First learn to convert all the units into SI units.Then you will be fine.(If you do it correctly)

 

[Medgirl314]

Sorry, I must be phrasing something wrong. I understand that I neglected to convert the units. I know how, I just forgot. What I'm really trying to ask here is if, in this particular case, it affected the answer numerically. I got the units wrong, but was the number part right? I'm failing to see where the units would have affected my numerical answer.

Thanks!

 

[adjacent]

Sorry, I must be phrasing something wrong. I understand that I neglected to convert the units. I know how, I just forgot. What I'm really trying to ask here is if, in this particular case, it affected the answer numerically. I got the units wrong, but was the number part right? I'm failing to see where the units would have affected my numerical answer.

Thanks!

What did you use for acceleration? 9.8m/s² ?
or did you convert that to feet?
What is the difference between 9.8meter/s² and 9.8feet/s² ?
It all depends on what you used for a

 

[Medgirl314]

I used meters. I think I see it now! Was the time wrong? I see that I used feet for y and meters for a, is that what is wrong with the answer?

 

[ehild]

9.81 m is not the same as 9.81 ft, is it? What length is one ft in meters?

When you use an equation, the units are together with the data. So you got the numerical result for time $$t=\sqrt{\frac{2y}{a}}$$ as

$$t=\sqrt{\frac{44 ft} {9.81ms^{-2 }}}=2.12 s \sqrt{\frac{ft}{m}}$$, in [s sqrt(ft/m)] units.
If you multiply it with the speed in ft/s, you get the distance in $\sqrt{\frac{ft^3}{m}}$.

Does it have sense?

ehild

 

[SteamKing]

Yes, units are not a style choice: they do affect the numerical values in calculations.

Hint: since you are working in feet, g is taken as 32.2 ft/s^2. You don't necessarily have to convert all your calculations if you don't want to.

 

[adjacent]

I used meters. I think I see it now! Was the time wrong? I see that I used feet for y and meters for a, is that what is wrong with the answer?

Lets see now.
You used the equation:y=1/2a(t^2)
So $t=\sqrt{\frac{2y}{a}}$ (ah your equation was wrong.You wrote $t=\frac{\sqrt{2}y}{a}$
If a=9.8m/s^2,
$t=\sqrt{\frac{2*22}{9.8}}$
=2.12s
We used feet for y and meters for a.This is wrong,right?
Then if we use feet instead of meters for a,
$t=\sqrt{\frac{2*22}{32.1522}}$
=1.17s
This answer is right because we used feet for both y and a.
If we use meters for both y and a,
$t=\sqrt{\frac{2*6.7056}{9.8}}$
=1.17s

The last two answers were same.
because we used the same units for y and a.

 

[Medgirl314]

Ehelid and SteamKing,thanks for the explanations. I know units aren't optional, I just didn't see *where* they came into play for this particular problem, so I was trying to achieve some level of clarification on that before moving on.

Thank you, adjacent, that was an incredibly helpful explanation. That's what I thought, I just didn't think about the "y" being in feet at first. Thanks for clarifying where I went wrong!

Okay, I know you already calculated the time, adjacent, but I went over it again anyway just to be sure I understood. Could someone please tell me where the square root button can be found?

I, too, got 1.17 seconds for the time.

Horizontal:
x=vt
x=250ft(1.17s)
x=293 ft

How's that?

 

[adjacent]

Could someone please tell me where the square root button can be found?

It can be found on the right side of the post of the Advanced editor.(Quick symbols)
You can also use LaTeX. At the top right side of the advanced editor

Horizontal:
x=vt
x=250ft(1.17s)
x=293 ft

How's that?

That's right

 

[Medgirl314]

Ah, found it! Thank you. That's it for this problem, correct? Thanks again!