# An outfielder throws a 0.150-kg baseball at a speed of 24.0 m/s and an initial angle of...

## Question:

An outfielder throws a 0.150-kg baseball at a speed of 24.0 m/s and an initial angle of 30.0{eq}^{\circ} {/eq}. What is the kinetic energy of the ball at the highest point of its motion?

## Kinetic Energy:

Let us assume a Dexter cattle of mass (m) is running on the ground with a speed (v). In this scenario, the kinetic energy of the given cattle would be expressed as,

{eq}\displaystyle K = \frac{1}{2}mv^{2}. {/eq}

The kinetic energy is expressed in Joules (J) in the SI unit system.

Given data:

• Mass of the baseball, {eq}m = 0.150 \ kg {/eq}
• Initial speed, {eq}v = 24.0 \ m/s {/eq}
• Projection angle, {eq}\theta = 30.0^{\circ} {/eq}

At the highest point, the vertical component of the velocity becomes zero. At this point, the baseball has only horizontal component of velocity which always remains unchanged during motion.

Thus, the kinetic energy of the baseball at this point can be given as,

{eq}\begin{align*} K &= \frac{1}{2}m v\textrm{cos}30^{\circ}\\ K &= \frac{1}{2} \times 0.150 \times 24.0 \times 0.866\\ K &= 1.56 \ \rm J.\\ \end{align*} {/eq}