A particle thrown at an angle theta with the vertical calculate maximum kinetic energy at highest point
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Hey
+_+ Considering the word 'Maximum', I assume we can certainly play with the angle 'θ' ...
Firstly,
--> K.E. ∝ v²
Also, keeping in mind -->
-> K.E. + P.E. = Constant [ Law of Conservation of Energy ]
We need the minimum vertical velocity, so that the velocity at highest point is not equal to zero ( such that the P.E. is zero )
Such a thing happens at 'θ' = 0°, when particle is at linear motion in one direction...
Hence, Maximum K.E. at highest point can be seen as ( 0.5 mv² ) when particle is at linear motion..
This can be mathematically expressed as :->
[tex]Vertical \ velocity = v cos \theta \\ \\ From \ third \ law \ of \ motion \ \ \ --\ \textgreater \ \\ \\ -- \ h = \frac{( v cos \theta )^2}{2g} \\ \\[/tex]
Now,
[tex]K.E. \ at \ peak = \frac{1}{2} m ( (vcos \theta )^2 + 2gh ) \\ \\ But, 2gh = ( vcos \theta )^2[/tex]
---------------------> which implies that our K.E. at maximum Height will always be zero
Hope this helped ^_^
+_+ Considering the word 'Maximum', I assume we can certainly play with the angle 'θ' ...
Firstly,
--> K.E. ∝ v²
Also, keeping in mind -->
-> K.E. + P.E. = Constant [ Law of Conservation of Energy ]
We need the minimum vertical velocity, so that the velocity at highest point is not equal to zero ( such that the P.E. is zero )
Such a thing happens at 'θ' = 0°, when particle is at linear motion in one direction...
Hence, Maximum K.E. at highest point can be seen as ( 0.5 mv² ) when particle is at linear motion..
This can be mathematically expressed as :->
[tex]Vertical \ velocity = v cos \theta \\ \\ From \ third \ law \ of \ motion \ \ \ --\ \textgreater \ \\ \\ -- \ h = \frac{( v cos \theta )^2}{2g} \\ \\[/tex]
Now,
[tex]K.E. \ at \ peak = \frac{1}{2} m ( (vcos \theta )^2 + 2gh ) \\ \\ But, 2gh = ( vcos \theta )^2[/tex]
---------------------> which implies that our K.E. at maximum Height will always be zero
Hope this helped ^_^
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