关于求抛射角的最佳值的题目,一再引起广大物理教师和学生的很大兴趣。自60年代开始至今,不少书刊不断登载了各种各样的解法,如:用微分的方法,用判别式的方法,用三角函数的方法,用不等式的方法,等等,然而步骤都较繁.现介绍一种简单方法。题目在掷铅球的运动中,铅球出手时距地面的高度为 h,若出手时的速度为 v_0,求以何角度掷球,水平射程最远?其值若干?解:在附图的矢量三角形中,根据正弦定理:gt/sin(α_1+α_2)=v_0/cosα_2 ①,水平射程:x=t·v_tcosα_2 ②,因为机械能守恒,∴v_t=(v_0~2+2gh)~(1/2) ③①×②×③:x=((v_0(v_0~2+2gh)~(1/2))/g)sin(α_1+α_2)④ The question of finding the best value of the projectile angle has caused great interest among the physics teachers and students. Since the beginning of the 1960s, many books and periodicals have continuously published various solutions, such as using differential methods, discriminative methods, trigonometric methods, inequality methods, etc. Fan. Now introduce a simple method. In the shot-throwing exercise, the height of shot put is h from the ground, if the speed is v_0, what angle is required to throw the ball, the horizontal range is farthest? Its value is several: Solution: The vector triangle in the figure According to the sine theorem: gt/sin(α_1+α_2)=v_0/cosα_2 1, horizontal range: x=t·v_tcosα_2 2. Since mechanical energy is conserved, ∴v_t=(v_0~2+2gh)~(1/2) 31×2×3:x=((v_0(2+0~2gh)~(1/2))/g)sin(α_1+α_2)4