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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

Sagot :

Trancescient

Explanation:

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

[tex]KE = \frac{1}{2}m(v_0\cos30)^2[/tex]

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

[tex]\;\;\;\;\;= 90\:\text{J}[/tex]

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