Projectile motion problem, given the KE at the top of the parabolic arc

AI Thread Summary
The discussion focuses on calculating the angle of projection where the kinetic energy at the highest point of a projectile's trajectory equals one-fourth of its kinetic energy at the point of projection. Initial attempts suggested an angle of 60 degrees, but further analysis revealed that the horizontal component of velocity must be considered in the total kinetic energy. The correct approach involves using the relationship between the sine and cosine functions to derive the angle, leading to an answer around 61 degrees, although some participants believe it should be 60 degrees. Clarifications were made regarding the placement of the 0.25 factor in the equations. The conversation emphasizes the importance of understanding both vertical and horizontal kinetic energy components in projectile motion calculations.
Bilal Rajab Abbasi
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Homework Statement



Calculate the angle of Projection for which Kinetic Energy at the highest point of trajectory equal to one-fourth of its kinetic energy at point of projection?

Homework Equations



Range and height of Projectile equations

The Attempt at a Solution


[/B]
I've made two equations k.e wise and velocity wise but the answer which is 60 degrees isn't quite right.
 
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Bilal Rajab Abbasi said:

Homework Statement



Calculate the angle of Projection for which Kinetic Energy at the highest point of trajectory equal to one-fourth of its kinetic energy at point of projection?

Homework Equations



Range and height of Projectile equations

The Attempt at a Solution


[/B]
I've made two equations k.e wise and velocity wise but the answer which is 60 degrees isn't quite right.
Please post your work in detail so we can check it. Thanks. :smile:
 
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0.5 * mv^2sinx=0.25*(1/2 mv^2cosx)
Cosx/sinx=4 after cutting the common values
Tanx=sinx/cosx
Tanx=1/4
X=14.1
That was my attempt
In order to get the answer right I did this
(tan inverse of 4 minus tan inverse of -1/4
Answer is 61 degrees neglecting negative angle
But actually its 60.
 
Bilal Rajab Abbasi said:
0.5 * mv^2sinx=0.25*(1/2 mv^2cosx)
But the initial KE includes both the horizontal and vertical components, no? It looks like you are taking the initial vertical KE to be 1/4 of the horizontal KE at the top... Maybe I'm not understanding what you wrote...
 
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Can I get a solution on that please
I have my sendups coming and this question is part of the syllabus
It's an important one.
 
@berkeman is correct. You are ignoring the horizontal component of the velocity for the total kinetic energy at the initial point.

Bilal Rajab Abbasi said:
Can I get a solution on that please
No. Providing solutions (and asking to be provided solutions) is against the forum rules, which you agreed to when you signed up.
 
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Bilal Rajab Abbasi said:
0.5 * mv^2sinx=0.25*(1/2 mv^2cosx)
.

I would say your 0.25 is in the wrong place apart from what was mentioned above about the initial KE.
 
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Thanks. I had corrected that and solved it.
 
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