Projectile motion hitting angle

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Homework Help Overview

The problem involves determining the minimum velocity required for a projectile to hit a target embedded in the side of a mountain at a distance of L meters, with the projectile launched at an angle of π/4. The context is projectile motion, focusing on the relationship between launch angle, initial velocity, and impact conditions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to write out equations of motion and identify relevant variables. There is uncertainty about the height of the target and how to support assumptions regarding it. The original poster expresses confusion about deriving the necessary expressions for velocity and impact angle.

Discussion Status

Participants are actively engaging with the problem, sharing equations and exploring the relationships between variables. Some guidance has been provided regarding the assumptions that can be made about height and the impact angle, and there is a focus on deriving expressions for velocities at impact.

Contextual Notes

There is a lack of explicit information about the height of the target, which is a point of contention in the discussion. Participants are considering the implications of assuming the height does not matter and how that affects the problem setup.

loops496
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Homework Statement



You're given a cannon that allows you to fix the initial velocity and the shooting angle of a projectile that should be embedded on the side of a mountain [itex]l[/itex] meters away.

What is the minimum velocity for you to embed the projectile at an angle of [itex]\frac{\pi}{4}[/itex]?

The Attempt at a Solution



I'm not quite sure how to tackle the problem, I do know I have to optimize somehow the velocity but I don't know which expression should I take the derivative of. All help will be greatly appreciated.
M.
 
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Always start by writing out the equations of motion, creating variables as necessary.
You don't give a height for the target. If none is given, what height (relative to the cannon) do you think you should assume?
Of the variables velocity, time, launch angle, x-position, y-position etc., which ones do you think are relevant? Eliminate the others to arrive at one equation with one variable, then optimise it.
 
Hey haruspex thanks for replying.
As you suggested I had indeed written down the equations of motion (I had to because it is a problem with multiple questions with the one I posted as the last one). The thing I'm Confused with is the one you note in your reply, I don't know what height to assume. I guess at the same level as the cannon but don't know how to support that answer. I think the only variables that play a role in the answer are shooting angle and initial velocity, and I also know that for the projectile to hit at [itex]\frac{\pi}{4}[/itex] the components of the velocity i.e. [itex]v_{fy}[/itex] & [itex]v_{fx}[/itex] right before the impact should be the same but still I don't know how to get an answer analytically.
 
loops496 said:
The thing I'm Confused with is the one you note in your reply, I don't know what height to assume. I guess at the same level as the cannon but don't know how to support that answer.
No, you need to assume the height doesn't matter. If you were given the range, the height and the arrival angle then you'd have no room for movement. The launch angle and speed would already be determined.
So, please post some equations. Let the launch be speed v angle theta. What does the range L give you for the flight time? What would the horizontal and vertical velocities be at impact?
 
Oh I get it!
Ok for flight time I got [tex]t_f = \frac{L}{v cos \theta}[/tex] as for velocities the x component is the same as the launch one i.e. [tex]v_{fx} = v cos \theta[/tex] and for the y component [tex]v_{fy} = v sin \theta - \frac{gL}{v cos \theta}[/tex]
 
loops496 said:
Oh I get it!
Ok for flight time I got [tex]t_f = \frac{L}{v cos \theta}[/tex] as for velocities the x component is the same as the launch one i.e. [tex]v_{fx} = v cos \theta[/tex] and for the y component [tex]v_{fy} = v sin \theta - \frac{gL}{v cos \theta}[/tex]

Right, so what is the impact angle from those?
 
The angle is then [tex]\theta_i = tan^{-1}\left( tan \theta - \frac{gL}{(v cos \theta)^2}\right)[/tex]
 
loops496 said:
The angle is then [tex]\theta_i = tan^{-1}\left( tan \theta - \frac{gL}{(v cos \theta)^2}\right)[/tex]
Yes, and that has to be equal to... ?
 
Obviously to [itex]\frac{\pi}{4}[/itex]. And then should I take the derivative with respect to v and optimize??
 
  • #10
loops496 said:
Obviously to [itex]\frac{\pi}{4}[/itex]. And then should I take the derivative with respect to v and optimize??

Yes, but I suggest going back to your vfx, vfy equations. How can you express a 45 degree downward impact angle in terms of a relationship between those velocities?
 
  • #11
I think something like [tex]\frac{v_{fy}}{v_{fx}} = -1[/tex] should do it, right?
 
  • #12
loops496 said:
I think something like [tex]\frac{v_{fy}}{v_{fx}} = -1[/tex] should do it, right?

Looks good.
 
  • #13
That's some quality help you gave me, I appreciate it!

Many thanks,

M.
 
  • #14
loops496 said:
That's some quality help you gave me, I appreciate it!

Many thanks,

M.

You're welcome. Thank you for an interesting problem.
 

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