標題:
physics ball
發問:
According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy "Dizzy" Carlyle in a minor league game. The ball traveled 188 m (618 ft) before landing on the ground outside the ballpark.Assuming the ball's initial velocity was 53∘ above the... 顯示更多 According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy “Dizzy” Carlyle in a minor league game. The ball traveled 188 m (618 ft) before landing on the ground outside the ballpark. Assuming the ball's initial velocity was 53∘ above the horizontal and ignoring air resistance, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point 0.9 m (3.0 ft) above ground level? Assume that the ground was perfectly flat. How far would the ball be above a fence 3.0 m (10 ft) high if the fence was 116 m (380 ft) from home plate?
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最佳解答:
Let U be the initial velocity of the ball. Consider the horizontal motion: 188 = (U.cos(53)).t where t is the time of travel of the ball hence, t = 188/(U.cos(53)) ------------- (1) Consider the vertical motion: use equation: s = ut + (1/2)at^2 with s = -0.9 m, u = U.sin(53), a = -g(=-10 m/s^2), t = 188/(U.cos(53)) hence, -0.9 = [U.sin(53)].[188/(U.cos(53))] + (1/2).(-10).[188/(U.cos(53))]^2 i.e. -0.9 = 188.tain(53) - 5.(188)^2/(U.cos(53))^2 solve for U gives U = 44.14 m/s The initial speed of the ball is 44.14 m/s Time of flight of the ball to the point above the fence t' is t' = 116/[44.14.cos(53)] s = 4.367 s Consider the vertical motion: Use equation: s = ut + (1/2)at^2 with u = 44.14sin(53) m/s = 36.59 m/s, t = 4.367 s, a = -g(= -10 m/s^2), s =? hence, s = [36.59 x 4.367 + (1/2)(-10).(4.367)^2] m = 64.43 m Thus, height above fence = (64.43 + 0.9 - 3) m = 62.34 m
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