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Question: A cricket ball is hit for a six leaving the bat at an angle of \({45^ \circ }\) to the horizontal wi...

A cricket ball is hit for a six leaving the bat at an angle of 45∘{45^ \circ } to the horizontal with kinetic energy KK. At the top, kinetic energy of the ball is:
A) 00

B) KK

C) K2\dfrac{K}{2}

D) K2\dfrac{K}{{\sqrt 2 }}

Explanation

Solution

While moving horizontally at maximum height, there will only be horizontal velocity and the vertical velocity will be zero.
Also we have to use the kinetic energy formula here.

Complete step by step solution:
We know that:
At the top of maximum height of any object there is only a horizontal component and the vertical component becomes zero.
So, velocity at the top will be:
vtop=vcos⁡45∘{v_{top}} = v\cos {45^ \circ }

We know that,
Kinetic energy,
K=12mv2 =12m(vcos⁡45∘)2 =12mv22 =K2  K = \dfrac{1}{2}m{v^2} \\\ = \dfrac{1}{2}m{\left( {v\cos {{45}^ \circ }} \right)^2} \\\ = \dfrac{{\dfrac{1}{2}m{v^2}}}{2} \\\ = \dfrac{K}{2} \\\

Hence, option C is the answer.

Additional information: Kinetic energy- The form of energy that an object or a particle has because of it motion is known as kinetic energy. If function which absorbs energy is performed on an object by applying net force, the object speeds up and thus receives kinetic energy.
Kinetic energy is the property of a moving medium or particle which depends not only on its speed, but also on its density. The form of motion can be translation, rotation of an axis or vibration.
Mathematically, the kinetic energy is given by-
K.E =12mv2 = \dfrac{1}{2}m{v^2}

In the kinetic energy equation we can see that kinetic energy is equal to the product of mass and square of velocity. A square of velocity and product of mass cannot be negative. So, kinetic energy cannot be negative.
Although kinetic energy is never negative, the change in kinetic energy may be negative. Kinetic energy is energy in motion, so an object in motion cannot have negative energy.

Note: Here while writing the velocity we have to pay attention to the fact that velocity will not be directly equal to cos⁡45∘\cos {45^ \circ }. Instead it will be equal to velocity times cos⁡45∘\cos {45^ \circ } as we are taking the horizontal component.