Projectile Motion - is this solvable?

Click For Summary

Homework Help Overview

The discussion revolves around a projectile motion problem involving an arrow shot at a 45-degree angle from a 51.2m cliff, landing at a 35-degree angle. Participants explore the feasibility of solving this problem, particularly in the context of a test question.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the initial attempts to solve the problem using tangent functions and question the appropriateness of treating it as a geometry problem rather than a physics problem. There are inquiries about the relevant equations and the physical implications of the projectile's path.

Discussion Status

The discussion is ongoing, with some participants expressing uncertainty about the problem's solvability. There are suggestions to reconsider the assumptions made in the problem setup, and some participants have provided insights into the equations of motion relevant to projectile motion.

Contextual Notes

There is mention of potential errors in the problem statement provided by the teacher, as well as varying interpretations of the angle at which the arrow lands. Some participants note that the problem may not have a solution under certain interpretations, while others suggest that it could be solvable with the right approach.

Jrslager

Homework Statement


An arrow is shot at 45 degrees into the sky off of a 51.2m cliff. It lands at a 35 degree angle

Homework Equations

The Attempt at a Solution


I attempted to solve it using tangent of 35 is equal to the height (51.2m) divided by the distance it traveled which equals 72. I more just need to know if this is solvable given it was on a test of mine and I would like to understand how to solve it for next time.
 
Last edited by a moderator:
Physics news on Phys.org
Jrslager said:

Homework Statement


An arrow is shot at 45 degrees into the sky off of a 51m cliff. It lands at a 35 degree angle

Homework Equations

The Attempt at a Solution


I attempted to solve it using tangent of 35 is equal to the height (51m) divided by the distance it traveled which equals 72. I more just need to know if this is solvable given it was on a test of mine and I would like to understand how to solve it for next time.
Welcome to the PF. :smile:

Seems solvable to me, but I haven't worked it out yet. Please post the Relevant Equations and show us your work in detail. A sketch of the problem would also help.
 
Jrslager said:
tangent of 35 is equal to the height (51m) divided by the distance it traveled
So you think it traveled in a straight line from the cliff edge to the ground?
 
Jrslager said:

Homework Statement


An arrow is shot at 45 degrees into the sky off of a 51m cliff. It lands at a 35 degree angle

Homework Equations

The Attempt at a Solution


I attempted to solve it using tangent of 35 is equal to the height (51m) divided by the distance it traveled which equals 72. I more just need to know if this is solvable given it was on a test of mine and I would like to understand how to solve it for next time.
Yes, it's solvable. From the description of your attempt, however, it sounds like you're not considering the physics of the situation at all. Instead, you're looking at it as a geometry problem. That was your first mistake, and unfortunately, it's a showstopper.

I suggest you review the examples in your textbook on projectile motion. Ask yourself, "Why don't they just use simple geometry to solve these examples?" If you already know the answer to that question, ask yourself, "Why did I try to solve this problem using simple geometry?" so you can avoid making the same mistake in the future.
 
  • Like
Likes   Reactions: cnh1995
I know it didnt travel at a straight line down, I just do not know how to figure out the height that the arrow travels to. I recognize that this is physics and not geometry, you guys asked for an attempt...
 
Jrslager said:
I know it didnt travel at a straight line down, I just do not know how to figure out the height that the arrow travels to. I recognize that this is physics and not geometry, you guys asked for an attempt...
It's good that you recognised your method was wrong.

You did not list any relevant equations. You must have been taught some relating to motion under constant acceleration. These are sometimes referred to as the SUVAT equations.
 
Just wanted to know if it is solvable (my teacher gives me harder problems than the rest of the class in order to challenge me but there were mistakes in the question ie he wrote car (rest of the class had the problem as a car drives straight off the cliff horizontally, given the initial height and impact angle calculate the rest of the variables) instead of arrow and i believed he may have messed up). My teacher gave me the problem with the 45 degree takeoff point rather than a horizontal takeoff point that the rest of the class got in order to challenge me I have found out. Got full credit on the problem by solving for the horizontal takeoff point but I still want to find out how to calculate speed and time when I am only given the angle that it leaves from, angle it finishes at, and height that the initial shot is taken off of.
 
Last edited by a moderator:
Physics eq.jpg

This is all that is given in the problem
 
Jrslager said:
Got full credit on the problem by solving for the horizontal takeoff point
So what standard equations did you use?
 
  • #10
X= XsubO + VsubO x t + 1/2 a t^2

V= VsubO + at

g= 9.8m x s^-1
 
  • #11
Jrslager said:
X= XsubO + VsubO x t + 1/2 a t^2

V= VsubO + at
Ok, but since this is two dimensional and the acceleration would be in the vertical direction, it would be more usual to use y instead of x there and reserve x for horizontal motion.
Suppose it was fired at speed v. With what vertical and horizontal speeds would it land?
 
  • #12
This is not for homework help... this was a test question that i wanted to know how to solve. I got full credit so i don't care anymore, thanks anyways
 
  • #13
Jrslager said:
This is not for homework help... this was a test question that i wanted to know how to solve. I got full credit so i don't care anymore, thanks anyways
Well, do you want to find out how to solve it?
 
  • #14
Yes, that has been what I have been asking this entire time :)
 
  • #15
Jrslager said:
This is not for homework help... this was a test question that i wanted to know how to solve. I got full credit so i don't care anymore, thanks anyways
Please re-read the PF rules about schoolwork (see INFO at the top of the page). All schoolwork-type questions go here in the Homeworkk Help forums, with the Template filled out with the Relevant Equations and your Attempt at the Solution.
 
Last edited:
  • #16
Jrslager said:
An arrow is shot at 45 degrees into the sky off of a 51.2m cliff. It lands at a 35 degree angle
Upon further review, I now think that this problem does not have a solution. Can you think of a reason why I would say that based on the projectile equations of motion?
Jrslager said:
I got full credit so i don't care anymore, thanks anyways
Can you post your solution? I'd like to see how you got full credit for solving this problem that has no solution...
 
Last edited:
  • #17
berkeman said:
Upon further review, I now think that this problem does not have a solution. Can you think of a reason that why I would say that based on the projectile equations of motion?
Hint -- What is non-physical about your sketch in post #8? (well, except for it being posted sideways...)
 
  • #18
Jrslager said:
Yes, that has been what I have been asking this entire time :)
Then please try to answer my question in post #8: Suppose it was fired at speed v. With what vertical and horizontal speeds would it land?
 
  • #19
berkeman said:
I now think that this problem does not have a solution.
I believe it does. See the note I sent you.

Edit: after seeing a note from @kuruman , perhaps I misunderstood your reason for saying it has no solution. I had assumed you meant there was insufficient information.
As kuruman notes, you have to assume the given 35 degrees is the angle to the vertical.
 
Last edited:
  • #20
I got full credit by solving the basic projectile motion problem of the arrow being shot horizontally rather than at a 45 degree angle and he accepted it given that the problem with the 45 degree shot is nearly impossible (if it is even possible)
 
  • #21
haruspex said:
I believe it does. See the note I sent you.

Edit: after seeing a note from @kuruman , perhaps I misunderstood your reason for saying it has no solution. I had assumed you meant there was insufficient information.
As kuruman notes, you have to assume the given 35 degrees is the angle to the vertical.
After the PM exchanges, I now see that there is actually a solution (not the 35 degrees to the vertical, though). Very clever and non-intuitive. Thanks. :smile:
 
Last edited:
  • #22
Would you mind elaborating? My teacher would like to see given that he says that there is no solution
 
  • #23
Jrslager said:
Would you mind elaborating? My teacher would like to see given that he says that there is no solution
Draw the parabola of the flight path of the arrow. It starts off at 45 degrees above the horizontal. Find the two spots on the parabola where the arrow forms a 35 degree angle to the horizontal...
 
  • #24
And note that the exact problem statement allows some latitude in the possible answer...
Jrslager said:
An arrow is shot at 45 degrees into the sky off of a 51.2m cliff. It lands at a 35 degree angle
 
  • #25
Wait, It would be physically impossible. Just realized that the arrow COULD NOT land at 35 degrees from horizontal if shot at 45 degrees vertical
 
  • #26
Jrslager said:
Wait, It would be physically impossible. Just realized that the arrow COULD NOT land at 35 degrees from horizontal if shot at 45 degrees vertical
Yes it can. Make the sketch that I described in my previous post.
 
  • #27
Jrslager said:
COULD NOT land at 35 degrees from horizontal
Not if landing below the launch point certainly. But the problem as you posted it does not specify that the 35° is to the horizontal.
 
  • #28
Perhaps the initial speed was rather large.
upload_2017-10-18_11-45-10.png
 

Attachments

  • upload_2017-10-18_11-45-10.png
    upload_2017-10-18_11-45-10.png
    12.3 KB · Views: 363
  • Like
Likes   Reactions: Jrslager, vela and donpacino
  • #29
Jrslager said:
View attachment 213230
This is all that is given in the problem

You should think about why your diagram shows an impossible situation. The arrow cannot land lower than the starting point but at a shallower angle from the launch angle. (That is, unless the arrow goes part way around the world, as shown in #28.)
 
  • #30
Ray Vickson said:
You should think about why your diagram shows an impossible situation. The arrow cannot land lower than the starting point but at a shallower angle from the launch angle. (That is, unless the arrow goes part way around the world, as shown in #28.)
I think Jrslager got to that in post #25.
 

Similar threads

Find the height of a lamp based on the velocities of projectiles
  • · Replies 40 ·
2
Replies
40
Views
3K
Find Projectile Flight Time Given Only Maximum Height
  • · Replies 4 ·
Replies
4
Views
3K
Projectile Motion - object that is shot at an angle of 55 degrees
  • · Replies 8 ·
Replies
8
Views
2K
I need opinions on a Projectile Motion problem that I made up
  • · Replies 11 ·
Replies
11
Views
2K
How Do You Solve a Projectile Motion Problem?
  • · Replies 3 ·
Replies
3
Views
4K
Relationship between Launch height and range of a projectile
  • · Replies 15 ·
Replies
15
Views
27K
Relative motion/Projectile motion
  • · Replies 18 ·
Replies
18
Views
5K
Projectile Motion Experimental Error
  • · Replies 1 ·
Replies
1
Views
2K
Projectile motion problem, given the KE at the top of the parabolic arc
  • · Replies 7 ·
Replies
7
Views
2K
Where on the hill does the projectile land?
  • · Replies 9 ·
Replies
9
Views
3K