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

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

$KE = \frac{1}{2}m(v_0\cos30)^2$

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

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