Question

A cricket ball is hit for a six leaving the bat at an angle of θ to the horizontal with kinetic energy ‘k’. At the top, K.E. of the ball is

Answer 1

At the top, K.E. of the ball is (1/2) mv²Cos²θ

Explanation:

K.E  is Kinetic Energy

Kinetic energy = (1/2) mv²

Kinetic energy at bottom  =  (1/2) mv² = k

Velocity has two components

Horizontal = v cosθ

Vertical = v sinθ

Kinetic energy at bottom  =  (1/2) mv²Cos²θ  + (1/2)mv²Sin²θ

at top vertical velocity becomes 0

Hence vsinθ  = 0  => v²Sin²θ = 0

Kinetic energy at top = (1/2) mv²Cos²θ  + 0

= (1/2) mv²Cos²θ

At the top, K.E. of the ball is (1/2) mv²Cos²θ

(1/2) mv²Cos²θ  has converted into PE  mgh

=> h  = v²Cos²θ/2g

Height of ball

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