A baseball outfielder throws a baseball of mass 0.24 kg at a speed of 48 m/s and initial angle of 28 degrees. What is the kinetic energy (in J) of the baseball at the highest point of the trajectory? Ignore air friction.
Solution)
Given,
Mass of baseball=0.24 kg
Initial speed=48 m/s
Angle, theta=28 degrees
Horuzintal component of velocity,
Vx=48 cos(28)
We know, at the highest point vertical velocity component=0
So,
Kinetic Energy=mVx/2=(.24 kg (48 cos(28))^2)/2=215.54 Joules (Ans)
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A baseball outfielder throws a baseball of mass 0.24 kg at a speed of 48 m/s...
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