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

[Bilal Rajab Abbasi]

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.

 

[berkeman]

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.

 

[Bilal Rajab Abbasi]

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.

 

[berkeman]

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...

 

[Bilal Rajab Abbasi]

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.

 

[Orodruin]

@berkeman is correct. You are ignoring the horizontal component of the velocity for the total kinetic energy at the initial point.

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.

 

[neilparker62]

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.

 

[Bilal Rajab Abbasi]

Thanks. I had corrected that and solved it.