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A body is projected at an angle of 45o with KE E. The K.E at the highest point is:
A. zero
B. 3E4
C. E2
D. E

Answer
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Hint: First of all we will find the value of velocity at the highest point of trajectory. Then, substitute that velocity value in the formula for kinetic energy. By further solving the equations we can derive the final answer.
Formula: K.E=12mv2

Complete answer:
When a body is projected at an angle of 45o it will follow a projectile motion.
We know that kinetic energy of a body is given by
K.E=12mv2
Where m represents mass of the body and v represents the velocity of the body.
As per our knowledge, we know the fact that at the highest point of the trajectory the vertical component of velocity becomes zero and the horizontal component is given by the formula ux=ucosθ
Where ‘u’ is the velocity with which the body is projected and angle given in this question is 45.
So let us now substitute θ=45o in horizontal component of velocity
So, we get ux=ucos45
ux=u2 as cos45=12
In the question it is given that body is projected with kinetic energy E that is
K.E=E=12mu2
Now, kinetic energy at the highest point will be
KE=12mux212m(u2)212×mu22E2
K.E=E2
 So, we get to the conclusion that at the highest point of trajectory the kinetic energy becomes half.

Therefore, Option (C) is correct.

Note:
An object thrown or projected into the air is subject to only the acceleration of gravity. The motion of the object under the acceleration of gravity is known as projectile motion. The object is called a projectile and the path followed by the object is called trajectory. The Range of the projectile is given by the formula R=u2sin2θg and the maximum height attained by it given by H=u2sin2θ2g.
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