Projectile motion hitting angle

In summary, the conversation discusses a problem where a cannon is given to embed a projectile on the side of a mountain at a certain angle and distance. The minimum velocity needed for the projectile to embed at an angle of \frac{\pi}{4} is being discussed and equations of motion are used to solve the problem. The relevant variables are identified and the impact angle is determined to be \frac{\pi}{4}. The conversation ends with the suggestion to take the derivative and optimize, and a thank you from M for the help.
  • #1
loops496
25
3

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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  • #2
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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
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]
 
  • #6
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?
 
  • #7
The angle is then [tex] \theta_i = tan^{-1}\left( tan \theta - \frac{gL}{(v cos \theta)^2}\right)[/tex]
 
  • #8
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... ?
 
  • #9
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.
 

Related to Projectile motion hitting angle

1. What is the angle of maximum range for a projectile?

The angle of maximum range for a projectile is 45 degrees. This is because at this angle, the horizontal and vertical components of the velocity are equal, resulting in the longest possible range.

2. How does the initial velocity affect the angle of a projectile's trajectory?

The initial velocity of a projectile does not affect the angle of its trajectory. The angle is determined solely by the direction in which it is launched.

3. What is the difference between the range and the horizontal displacement of a projectile?

The range of a projectile refers to the total distance it travels horizontally before returning to its initial height. The horizontal displacement, on the other hand, is the distance between the initial and final horizontal positions of the projectile.

4. How does air resistance affect the angle of a projectile's trajectory?

The presence of air resistance can cause a projectile to follow a curved path rather than a parabolic one. This can cause the angle of the trajectory to deviate from the expected angle calculated using the laws of projectile motion.

5. Can the angle of a projectile's trajectory be negative?

Yes, the angle of a projectile's trajectory can be negative. This indicates that the projectile is traveling in the opposite direction of the positive x-axis.

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