Question

A baseball outfielder throws a 0.150-kg baseball at a speed of 37.2 m/s and an initial angle of 26.0° to the horizontal. What is the kinetic energy of the baseball at the highest point of its trajectory?

Answer 1

K=83.84J

Step-by-step explanation:

The kinetic energy is calculated it the formula:

`K=(mv^2)/(2)`

The velocity at the highest point is only the horizontal component of the velocity (since there the vertical component is null), which is the same through all the trajectory, and is:

`v=v_0cos(\theta)`

Putting all together:

`K=(mv_0^2cos^2(\theta))/(2)=((0.15kg)(37.2m/s)^2cos^2(26^(\circ)))/(2)=83.84J`