Question

a cricket ball is hit at 60 degree horizontally with a kinetic energy k. what is the kinetic energy of the ball at the highest point?
a 0
b $\frac{k}{4}$
c $\frac{k}{2}$
d $\frac{k}{sqrt{2}}$

Explanation:

Step1: Recall kinetic - energy formula

The initial kinetic energy $K=\frac{1}{2}mv^{2}$. Let the initial velocity be $v$. The initial velocity components are $v_x = v\cos60^{\circ}$ and $v_y=v\sin60^{\circ}$.

Step2: Analyze velocity at highest - point

At the highest point of the projectile motion, the vertical component of velocity $v_y = 0$ and the horizontal component of velocity remains constant throughout the motion, $v_{x - final}=v_x=v\cos60^{\circ}=\frac{v}{2}$.

Step3: Calculate kinetic energy at highest - point

The kinetic energy at the highest point $K_{highest}=\frac{1}{2}mv_{x - final}^{2}$. Since $v_{x - final}=\frac{v}{2}$, then $K_{highest}=\frac{1}{2}m(\frac{v}{2})^{2}=\frac{1}{4}\times\frac{1}{2}mv^{2}$. And since $K = \frac{1}{2}mv^{2}$, we have $K_{highest}=\frac{K}{4}$.

Answer:

B. $\frac{K}{4}$