A cricket ball is hit for a six leaving the bat at an angle of 45^∘ to the horizontal with kinet

step 1 In this question we have given that a cricket ball is hit for a 6 that is leaving at a bat at th

A cricket ball is hit for a six leaving the bat at an angle...

A cricket ball is hit for a six leaving the bat at an angle of $45^{\circ}$ to the horizontal with kinetic energy $K$. At the top position, the kinetic energy of the ball is
(A) Zero
(B) $K$
(C) $K / 2$
(D) $K / \sqrt{2}$

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Key Concepts

Velocity Components

When an object is projected at an angle, its initial velocity splits into horizontal and vertical components. In the special case of a 45° launch, these components are equal in magnitude. However, at the highest point of its trajectory, the vertical component becomes zero due to gravitational deceleration, leaving only the horizontal component.

Kinetic Energy

Kinetic energy is the energy associated with the motion of an object, calculated as one-half the mass times the square of its speed (KE = 1/2 mv²). Since at the peak of a projectile’s trajectory the vertical speed is zero, the kinetic energy is solely due to the horizontal component of the velocity, which is half of the total kinetic energy at launch when the projectile is launched at 45°.

Projectile Motion

Projectile motion involves the two-dimensional movement of an object under the influence of gravity, where its horizontal and vertical motions are independent. At launch, the object's velocity is decomposed into horizontal and vertical components, which behave differently—horizontal motion remains constant (ignoring air resistance), while vertical motion changes due to gravity.

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