Asked by Anonymous
A pitched ball was hit by a batter at a 30ᵒ angle at a velocity of 40 m/s. The outfield wall, being very short, was just the same height as the swing of the batter’s bat. If the hit was just enough to have been a homerun, how far from the batter was the outfield wall?
Answers
Answered by
Henry
Dx = Vo^2*sin(2A)/g.
Dx = 40^2*sin(60)/9.8 = = 141.4 m.
Dx = 40^2*sin(60)/9.8 = = 141.4 m.
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