a cricket ball is it at angle 60 degree with the horizontal with kinetic energy K what will be the kinetic energy at the highest point of its trajectory
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Hi..
At the highest point of motion ,the vertical component of velocity becomes zero. Only horizontal component of velocity exists and it remains constant through out the motion(if air resistance is neglected).
As kinetic energy is given, it's velocity can be easily derived.
E = (1/2)×M×v^2.
=> V= √(2E/M).
The horizontal component of velocity will be(given angle of projection 60°) vcos60°.
=> V (horizontal) = √(2E/M)×(1/2) = √(E/M×1/2).
=> K.E at heighest point = 1/2×M× v ^2 .( Here v is horizontal velocity).
=> K.E = E/4...
Hope this helps u!!
At the highest point of motion ,the vertical component of velocity becomes zero. Only horizontal component of velocity exists and it remains constant through out the motion(if air resistance is neglected).
As kinetic energy is given, it's velocity can be easily derived.
E = (1/2)×M×v^2.
=> V= √(2E/M).
The horizontal component of velocity will be(given angle of projection 60°) vcos60°.
=> V (horizontal) = √(2E/M)×(1/2) = √(E/M×1/2).
=> K.E at heighest point = 1/2×M× v ^2 .( Here v is horizontal velocity).
=> K.E = E/4...
Hope this helps u!!
Rosedowson:
Pls mark d brainliest!!
what is E in this solution?
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sry
What is it .??
E is .actually its KE only
Pls mark d brainliest!!
Thanks..
welcome
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