Question

Baseball outfielder throws a baseball of mass 0.15 kg at a speed of 40 m/s and initial angle of 30. What

is the kinetic energy of the baseball at the highest point of the trajectory? Answer: 90 J

Answer 1

Step-by-step explanation:

At the top of the the ball's trajectory, there is only the horizontal component of the initial velocity, which is
`v_0\cos30,` so the kinetic energy of the ball at this point is

`KE = (1)/(2)m(v_0\cos30)^2`

`\;\;\;\;\;= (1)/(2)(0.15\:\text{kg})[(40\:\text{m/s})\cos30]^2`

`\;\;\;\;\;= 90\:\text{J}`