Projectile Motion find the highest angle

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Homework Help Overview

The discussion revolves around a projectile motion problem where the original poster attempts to determine the highest angle at which a projectile can clear a peak. The problem involves analyzing the motion of the projectile and its trajectory in relation to a specified height.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore different interpretations of the projectile's peak and the conditions under which it clears the mountain top. There are attempts to derive equations relating time, angle, and height, but some participants express confusion about the assumptions being made, particularly regarding the vertical velocity at the peak.

Discussion Status

There is an active exchange of ideas, with some participants suggesting alternative methods to approach the problem. Guidance has been offered regarding the relationship between the projectile's trajectory and the peak height, but no consensus has been reached on the correct interpretation or method to solve the problem.

Contextual Notes

Participants are grappling with the implications of the projectile's trajectory and the timing of reaching the peak in relation to the mountain height. There is an acknowledgment of the complexity of the problem, and some constraints related to the assumptions about vertical velocity at the peak are being questioned.

kukumaluboy
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Homework Statement



http://postimg.org/image/7yfw1zpq1/

http://postimg.org/image/7yfw1zpq1/ <-Question

The Attempt at a Solution


I don't really understand the question but ill try.

I need to first find the highest angle at which the projectile will clear the peak.
At Peak:

Sy = 0.5(u+Vpeak)t , vpeak = 0
1800 = 125tsinx
tsinx = 14.4
t = 14.4/sinx -equation1

Sx = 0.5+(Ux+Vpeakx)t , Ux=Vx i guess?
2500 = 0.5(250cosx)t
2500 = 125tcosx
tcosx = 20
t = 20/cosx
Substitute eq 2 into eq 1:
20/cosx = 14.4/sinx
cosx = 1.39sinx

Then i don't know how to solve
 
Last edited by a moderator:
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You need to rethink a bit. It is not necessary that the vertical velocity is zero when passing the peak. Try finding the height as a function of the horizontal distance for an arbitrary angle.
 
need more tips lol
 
This is a nice problem. I agree with Orodruin, that is probably the easiest method. What is your difficulty in following it?
 
At top of parabola Vy = 0

Vy = Uy + at
0 = 250sinx + 9.81t
t = 25.5sinx

After that?
 
kukumaluboy said:
At top of parabola Vy = 0

Yes, but the top of the parabola is not necessarily where the projectile passes the mountain top.
 
Orodruin said:
Yes, but the top of the parabola is not necessarily where the projectile passes the mountain top.

The top of the trajectory of the projectile can be reached before the peak is encountered, at the peak, or after the peak is encountered, so long as the altitude of the projectile clears the top of the peak.
 
SteamKing said:
The top of the trajectory of the projectile can be reached before the peak is encountered, at the peak, or after the peak is encountered, so long as the altitude of the projectile clears the top of the peak.

How does this contradict what I just said?
 

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