Final answer:
The kinetic energy at the highest point of a cricket ball hit at 30° will be solely due to its horizontal component of velocity, as the vertical component will be zero. Using the cosine of 30°, which is √3/2, the remaining kinetic energy is 3K/4. Therefore, the correct answer is (C) √3E/4.
Step-by-step explanation:
We are asked to find the kinetic energy at the highest point of the cricket ball's trajectory when it is hit at a 30° angle with the horizontal with kinetic energy K. In projectile motion, only the vertical component of the velocity changes due to gravity, while the horizontal component remains constant assuming no air resistance.
At the highest point of its trajectory, the vertical component of the cricket ball's velocity is zero, since gravity has slowed it down to a momentary stop before it starts to fall back to the ground. Therefore, the only kinetic energy the ball has at this point is due to its horizontal velocity.
Let us assume the initial velocity of the ball is v, and it can be broken down into two components: vx (horizontal) and vy (vertical). The kinetic energy K is given by ½mv2, where m is the mass of the ball.
Since the ball is hit at a 30° angle, vx = vcos(30°) and vy = vsin(30°). The horizontal kinetic energy at the highest point is therefore ½m(vcos(30°))2. Cos(30°) equals √3/2, so the kinetic energy is ½mv2(√3/2)2 = ½mv2(3/4) = ½K(3/4) = 3K/4.
The correct answer is then (C) √3E/4.