Projectile Motion - is this solvable?

[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.

 

[berkeman]

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.

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.

 

[haruspex]

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?

 

[vela]

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.

 

[Jrslager]

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...

 

[haruspex]

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.

 

[Jrslager]

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.

 

[Jrslager]

This is all that is given in the problem

 

[haruspex]

Got full credit on the problem by solving for the horizontal takeoff point

So what standard equations did you use?

 

[Jrslager]

X= XsubO + VsubO x t + 1/2 a t^2

V= VsubO + at

g= 9.8m x s^-1

 

[haruspex]

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?

 

[Jrslager]

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

 

[haruspex]

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?

 

[Jrslager]

Yes, that has been what I have been asking this entire time :)

 

[berkeman]

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.

 

[berkeman]

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?

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...

 

[berkeman]

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...)

 

[haruspex]

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?

 

[haruspex]

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.

 

[Jrslager]

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)

 

[berkeman]

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.

 

[Jrslager]

Would you mind elaborating? My teacher would like to see given that he says that there is no solution

 

[berkeman]

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...

 

[berkeman]

And note that the exact problem statement allows some latitude in the possible answer...

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

 

[Jrslager]

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

 

[berkeman]

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.

 

[haruspex]

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.

 

[TSny]

Perhaps the initial speed was rather large.

 

[Ray Vickson]

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.)

 

[haruspex]

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.