Archery and Projectiles-Honors

[Medgirl314]

Homework Statement

An arrow is fired from the top of a 60 m high cliff. It is fired at 80 m/s at a 30 degree angle above horizontal.

a)Find the time of the arrow's flight from start to impact.
b)Find the horizontal distance the arrow travels.
c)Find the velocity of the arrow(magnitude and direction) when it strikes the ground.


2. Homework Equations
sine
cosine
x=vt
t=(v-v0)/a

3. The Attempt at a Solution

a)
I calculated the vertical and horizontal velocities.
Horizontal: 80 cos 30= 69.8 m/s
Vertical:80 sine 30= 40 m/s

Then I calculated the time.

v0=40 m/s
a=-9.8 m/s^2
v=0
t=(v-v0)/a
t=(0-40)/-9.8
t=4.1

This was considering the time from start to peak, so we double to get 8.2 s.

b) x=vt
x=69.28(58.2)
x=4032 m.

This seems conceivable, but not exactly right.

c) How do I find the magnitude and direction? It is lost on me at the moment. If the arrow is *on* the ground, it's not moving, and so the velocity would be 0 m/s. But I am assuming the question means the velocity at the exact moment it hits the ground.

We aren't taking air resistance into account.

Thanks in advance!


I posted this problem previously, and was told to try this equation for the time instead: y=y0+v0t+1/2a*(t^2).

I did, and got this far:

2y=2y0+2v0t+a*t^2

120=0+2(69.8 m/s*t)+9.8 m/s^2(t^2)


But I suppose the other users grew tired of helping me and left. Would someone mind picking it back up? I have worked for at least an hour now on these problems, and I get the concept, but I do need help with some of the steps.

Thanks so much!

 

[SteamKing]

Your equation for time is a quadratic in t. Can you solve it?

 

[Medgirl314]

Oh, I thought you left! Yes, but I need a bit of help walking through it as I haven't done quadratics for quite awhile.

Is a=9.8 m/s^2,
b=2v0t, and
c=120 ?

Or do I need to rearrange the equation first?

 

[SteamKing]

Yes, you need to dust off that old devil Algebra and re-arrange the equation.

 

[Medgirl314]

Ah, okay. I don't mind algebra, it's somewhat fun.

120=0+2(69.8 m/s*t)+9.8 m/s^2(t^2)

We even already have a zero in it.
-9.8 m/s^2(t^2)+2(69.8 m/s*t)+120=0

How's that so far?

 

[SteamKing]

Keep going.

 

[Medgirl314]

Good, thanks!

-9.8 m/s^2(t^2)+2(69.8 m/s*t)+120=0

I'm not sure what to do next. Do I just make the replace the "x" in the quadratic formula with our "t" and use the formula? Or is there more simplification that should be done here?I don't know how to show the quadratic formula on here, so I guess I'll just have to give you my answer.Will that work?

 

[Medgirl314]

Also, are you still working on this thread: https://www.physicsforums.com/showthread.php?t=730642 ?

If not, I need to repost it so I can get a confirmation.

 

[haruspex]

Good, thanks!

-9.8 m/s^2(t^2)+2(69.8 m/s*t)+120=0

I'm not sure what to do next. Do I just make the replace the "x" in the quadratic formula with our "t" and use the formula? Or is there more simplification that should be done here?


I don't know how to show the quadratic formula on here, so I guess I'll just have to give you my answer.Will that work?

As a regular on this forum, maybe you should add LaTex to your skills.
$ax^2+bx+c = 0$
$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
Click Quote to see how that was written in LaTex. Note the use of { and } to group terms. So \frac{numerator}{denominator} gives you a fraction.
It's always a good idea to use the Preview button to see that you've got it right.

 

[Medgirl314]

Okay, thank you! I will try LaTex. Would you mind responding to the math part of that question? Can I already use the formula, or does more simplification need to be done?

Does a=9.8 m/s^2(t^2)
b=2(69.8 m/s*t)
and c=120 ?

I'm fairly sure about "c", the others I'm not positive.

Thank you again!

 

[haruspex]

Vertical:80 sine 30= 40 m/s

Then I calculated the time.

v0=40 m/s
a=-9.8 m/s^2
v=0
t=(v-v0)/a
t=(0-40)/-9.8
t=4.1

This was considering the time from start to peak, so we double to get 8.2 s.

The arrow was fired from ... where exactly? You want the time to impact. What do you know about where the impact will be?

c) How do I find the magnitude and direction? It is lost on me at the moment. If the arrow is *on* the ground, it's not moving, and so the velocity would be 0 m/s. But I am assuming the question means the velocity at the exact moment it hits the ground.

It says "when it strikes the ground", which I take to mean "with which it strikes the ground".

 

[Medgirl314]

The arrow was fired from the top of a 60 m high cliff. Are you saying that I need to include 60 m somewhere in my calculation? I'm not sure what we know about where the impact will be, other than it will be on the ground. Could you clarify the next step a little more? I'm not really sure whether to continue with the quadratic formula or use a different formula, since suggestions to change my calculations seem to have been made with your post.

Thanks for the comment about the velocity. Maybe you will help me with that part of the problem after we get the first part?

 

[haruspex]

The arrow was fired from the top of a 60 m high cliff. Are you saying that I need to include 60 m somewhere in my calculation?

Yes indeed. It is not entirely clear from the question, but I think you should assume the archer was facing out over the edge of the cliff, so the landing point will be 60m below the launch point. It's just a matter of inserting the right initial and final heights in the SUVAT equation.
This is instead of inserting a value for the final velocity - you do not know what that is, yet. SUVAT equations only apply while the acceleration is constant, so they will take you up to the instant before landing, but no further. It certainly won't be zero.

 

[Medgirl314]

Sorry about the delay! I'm a bit lost. Are we trying a new equation or continuing with the quadratic one, only changing it?

 

[lightgrav]

if you want to do it withOUT a quadratic, you do in order:
1) find the time to the peak
2) find the height to the peak (from launch, then add the cliff)
3) find the time from peak to valley-floor.
turns out it is exactly the same amount of effort as using the quadratic formula

but caution: in post 5 you inserted the horizontal velocity into the vertical location equation.