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I throw .150kg baseball at a speed of 40m/s at an angle of 30 degrees. What is the KE at the top of arc?
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- Iby KLv 71 decade agoFavorite Answer
at the top of the arch vertical is zero (converted to potential energy) so you only have horizontal component. note that horizontal velocity is constant while baseball is airborne (if we neglect air resistance, since there is nothing else acting on ball in horizontal axis).
Vx= 40m/s * cos(30)
Ek=0.5*m*Vx^2=0.5*0.150*[40*cos(3)]^2
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