A baseball outfielder throws a baseball of mass 0.36 kg at a speed of 34.0 m/s and initial angle of 30°.What is the kinetic energy of the baseball at the highest point of the trajectory? Ignore air friction.

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Quizplus · Accepted Answer

At the highest point of the trajectory, the vertical component of the velocity is zero, but the horizontal component remains unchanged. The kinetic energy (KE) at the highest point is due to the horizontal component of the velocity only. The initial velocity can be broken down into horizontal (v_x) and vertical (v_y) components using trigonometry. Given the angle (θ) is 30° and the speed (v) is 34.0 m/s:$$v_x = v \cdot \cos(\theta) = 34.0 \cdot \cos(30°) = 34.0 \cdot \frac{\sqrt{3}}{2} \approx 29.4 \, \text{m/s}$$The kinetic energy is given by the formula:$$KE = \frac{1}{2}mv^2$$Substituting the mass (m = 0.36 kg) and the horizontal velocity (v_x ≈ 29.4 m/s) into the formula gives:$$KE = \frac{1}{2} \cdot 0.36 \cdot (29.4)^2 \approx 156 \, \text{J}$$Therefore, the kinetic energy of the baseball at the highest point of its trajectory is approximately 156 J.

A Baseball Outfielder Throws a Baseball of Mass 0