一道英语物理题A quarterback passes the football downfield at 20m/s
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一道英语物理题
A quarterback passes the football downfield at 20m/s .It leaves his hand 1.8m above the ground and is caught by a receiver 30m away at the same height
What is the maximum height of the ball on its way to the receiver?
A quarterback passes the football downfield at 20m/s .It leaves his hand 1.8m above the ground and is caught by a receiver 30m away at the same height
What is the maximum height of the ball on its way to the receiver?
![一道英语物理题A quarterback passes the football downfield at 20m/s](/uploads/image/z/463748-68-8.jpg?t=%E4%B8%80%E9%81%93%E8%8B%B1%E8%AF%AD%E7%89%A9%E7%90%86%E9%A2%98A+quarterback+passes+the+football+downfield+at+20m%2Fs)
翻译如下:
一个球员以20m/s的速度抛出了一只足球,球离开他手时离地1.8m,接着,球在同一高度下被30米外的另一个球员接住了
请问,球的运动轨迹上可能的最大高度为多少?
要设角度来做的,设抛出时速度方向和水平夹角为A
所以球的水平速度是20cosA,竖直速度是20sinA,方向向上
所以球运动的时间就是t=s/v=30/20cosA
所以竖直速度就要满足20sinA/g=0.5*30/20cosA
算出角度A为0.5arcsin0.75,得出水平速度,竖直速度,运动时间
所以最大高度就为0.5*竖直速度*运动时间=3.386m
一个球员以20m/s的速度抛出了一只足球,球离开他手时离地1.8m,接着,球在同一高度下被30米外的另一个球员接住了
请问,球的运动轨迹上可能的最大高度为多少?
要设角度来做的,设抛出时速度方向和水平夹角为A
所以球的水平速度是20cosA,竖直速度是20sinA,方向向上
所以球运动的时间就是t=s/v=30/20cosA
所以竖直速度就要满足20sinA/g=0.5*30/20cosA
算出角度A为0.5arcsin0.75,得出水平速度,竖直速度,运动时间
所以最大高度就为0.5*竖直速度*运动时间=3.386m
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