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

In summary: But it looks like I made a mistake somewhere.In summary, the equation for angle of projection is: k.e. = 0.25*(1/2 mv^2cosx)
  • #1
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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  • #2
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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  • #3
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.
 
  • #4
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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  • #5
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.
 
  • #6
@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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  • #7
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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  • #8
Thanks. I had corrected that and solved it.
 
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1. What is projectile motion?

Projectile motion is the motion of an object through the air under the influence of gravity. It follows a curved path called a parabola.

2. How is the kinetic energy (KE) calculated at the top of the parabolic arc?

The kinetic energy at the top of the parabolic arc is calculated using the equation KE = 1/2 * m * v^2, where m is the mass of the object and v is the velocity at the top of the arc.

3. What factors affect the KE at the top of the parabolic arc?

The KE at the top of the parabolic arc is affected by the mass and velocity of the object, as well as the angle at which it is launched and any external forces acting on it (such as air resistance).

4. How is the KE related to the speed of the object?

The KE of an object is directly proportional to its speed. This means that as the speed of the object increases, the KE also increases.

5. What is the significance of knowing the KE at the top of the parabolic arc?

Knowing the KE at the top of the parabolic arc can help us understand how much energy the object has at that point, and how it will continue to move along its path. It also allows us to calculate other important variables, such as the maximum height and range of the object's motion.

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