a shot is fired at an angle of 60 degree to the horizontal with kinetic energy E. if air resistance is ignored, the kinetic energy at the top of the trajectory is:
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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.
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.
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